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Maths MCQ

Circle · Maharashtra Board STD 11 Science (MSBSHSE - English Medium). 158 questions.

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158 questions · 150 auto-graded MCQ + 8 self-marked written.

MCQ 11 Mark
The parametric equations of the circle $x^2+y^2+m x+m y=0$ are
  • ✓
    $x=\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta_i y=\frac{-m}{2}+\frac{m}{\sqrt{2}} \sin \theta$
  • B
    $x=\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta_2 y=\frac{+m}{2}+\frac{m}{\sqrt{2}} \sin \theta$
  • C
    x = 0, y = 0
  • D
    x = m cos θ, y = m sin θ
Answer
Correct option: A.
$x=\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta_i y=\frac{-m}{2}+\frac{m}{\sqrt{2}} \sin \theta$
$x=\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta_i y=\frac{-m}{2}+\frac{m}{\sqrt{2}} \sin \theta$
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MCQ 21 Mark
A pair of tangents are drawn to a unit circle with centre at the origin and these tangents intersect at A enclosing an angle of 60. The area enclosed by these tangents and the arc of the circle is
  • A
    $\frac{2}{\sqrt{3}}-\frac{\pi}{6}$
  • ✓
    $\sqrt{3}-\frac{\pi}{3}$
  • C
    $\frac{\pi}{3}-\frac{\sqrt{3}}{6}$
  • D
    $\sqrt{3}\left(1-\frac{\pi}{6}\right)$
Answer
Correct option: B.
$\sqrt{3}-\frac{\pi}{3}$
$\sqrt{3}-\frac{\pi}{3}$

Image

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MCQ 31 Mark
The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
  • A
    $x^2+y^2=9 a^2$
  • B
    $x^2+y^2=16 a^2$
  • ✓
    $x^2+y^2=4 a^2$
  • D
    $x^2+y^2=a^2$
Answer
Correct option: C.
$x^2+y^2=4 a^2$
$x^2+y^2=4 a^2$

Since the triangle is equilateral.

The centroid of the triangle is same as the circumcentre

and radius of the circumcircle $=\frac{2}{3}$ (median) $=\frac{2}{3}(3 a)=2 a$

Hence, the equation of the circumcircle whose centre is at $(0,0)$ and radius $2 \mathrm{a}$ is $\mathrm{x}^2+\mathrm{y}^2=$4a^2

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MCQ 41 Mark
If a circle passes through the points (0, 0), (a, 0), and (0, b), then find the co-ordinates of its centre.
  • A
    $\left(\frac{-a}{2}, \frac{-b}{2}\right)$
  • B
    $\left(\frac{a}{2}, \frac{-b}{2}\right)$
  • C
    $\left(\frac{-a}{2}, \frac{b}{2}\right)$
  • ✓
    $\left(\frac{a}{2}, \frac{b}{2}\right)$
Answer
Correct option: D.
$\left(\frac{a}{2}, \frac{b}{2}\right)$
$\left(\frac{a}{2}, \frac{b}{2}\right)$
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MCQ 51 Mark
The area of the circle having centre at (1, 2) and passing through (4, 6) is
  • A
    5π
  • B
    10π
  • ✓
    25π
  • D
    100π
Answer
Correct option: C.
25π
25π
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MCQ 61 Mark
If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle.
  • ✓
    $\frac{3}{4}$
  • B
    $\frac{4}{3}$
  • C
    $\frac{1}{4}$
  • D
    $\frac{7}{4}$
Answer
Correct option: A.
$\frac{3}{4}$
$\frac{3}{4}$

Tangents are parallel to each other.

The perpendicular distance between tangents = diameter

Image

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MCQ 71 Mark
The equation(s) of the tangent(s) to the circle $x^2+y^2=4$ which are parallel to $x+2 y+3=$ 0 are
  • A
    x – 2y = 2
  • B
    x + 2y = ±2√3
  • ✓
    x + 2y = ±2√5
  • D
    x – 2y = ±2√5
Answer
Correct option: C.
x + 2y = ±2√5
x + 2y = ±2√5
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MCQ 81 Mark
Find the equation of the circle which passes through the points (2, 3) and (4, 5), and the center lies on the straight line y – 4x + 3 = 0.
  • ✓
    $x^2+y^2-4 x-10 y+25=0$
  • B
    $x^2+y^2-4 x-10 y-25=0$
  • C
    $x^2+y^2-4 x+10 y-25=0$
  • D
    $x^2+y^2+4 x-10 y+25=0$
Answer
Correct option: A.
$x^2+y^2-4 x-10 y+25=0$
$x^2+y^2-4 x-10 y+25=0$
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MCQ 91 Mark
If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 sq. units, then find the equation of the circle.
  • A
    $x^2+y^2-2 x+2 y=40$
  • B
    $x^2+y^2-2 x-2 y=47$
  • ✓
    $x^2+y^2-2 x+2 y=47$
  • D
    $x^2+y^2-2 x-2 y=40$
Answer
Correct option: C.
$x^2+y^2-2 x+2 y=47$
$x^2+y^2-2 x+2 y=47$

Centre of circle = Point of intersection of diameters.

Solving 2x – 3y = 5 and 3x – 4y = 7, we get

x = 1, y = -1

Centre of the circle C(h, k) = C(1, -1)

∴ Area = 154

$\begin{aligned} & \pi r^2=154 \\ & \frac{22}{7} \times r^2=154 \\ & r^2=154 \times \frac{22}{7}=49 \\ & \therefore r=7\end{aligned}$

equation of the circle is

$\begin{aligned} & (x-1)^2+(y+1)^2=72 \\ & x^2+y^2-2 x+2 y=47\end{aligned}$

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MCQ 101 Mark
Equation of a circle which passes through (3, 6) and touches the axes is
  • A
    $x^2+y^2+6 x+6 y+3=0$
  • B
    $x^2+y^2-6 x-6 y-9=0$
  • ✓
    $x^2+y^2-6 x-6 y+9=0$
  • D
    $x^2+y^2-6 x+6 y-3=0$
Answer
Correct option: C.
$x^2+y^2-6 x-6 y+9=0$
$x^2+y^2-6 x-6 y+9=0$
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MCQ 112 Marks
Two tangents to the circle $x^2+y^2=4$ at the points $A$ and $B$ meet at the point $P(-4,0)$. Then the area of the quadrilateral $P A O B, O$ being the origin, is
  • A
    $2 \sqrt{3}$ sq. units
  • B
    $8 \sqrt{3} sq$. units
  • ✓
    $4 \sqrt{3} sq$. units
  • D
    $6 \sqrt{3} sq$. units
Answer
Correct option: C.
$4 \sqrt{3} sq$. units
(c) : The centre of the circle is $(0,0)$ and radius is 2 units and coordinates of $P$ are $(-4,0)$.
$\therefore O P=4$ units
Also $O A=O B=2$ units [radius of circle]
$
\therefore \quad O A^2+A P^2=O P^2
$
[By Pythagoras theorem]
$\Rightarrow 2^2+A P^2=(4)^2 \Rightarrow A P^2=12 \Rightarrow A P=2 \sqrt{3}$
Similarly, $B P=2 \sqrt{3}$
Area of $P A O B=$ Area of $\triangle O A P+$ Area of $\triangle O B P$
$=\frac{1}{2} \times 2 \sqrt{3} \times 2+\frac{1}{2} \times 2 \sqrt{3} \times 2=4 \sqrt{3}$ sq. units
Image
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MCQ 122 Marks
The centre of the circle whose radius is 3 units and touching internally the circle $x^2+y^2-4 x-6 y-12=$ 0 at the point $(-1,-1)$ is
  • ✓
    $\left(\frac{4}{5}, \frac{7}{5}\right)$
  • B
    $\left(\frac{4}{5}, \frac{-7}{5}\right)$
  • C
    $\left(\frac{-4}{5}, \frac{-7}{5}\right)$
  • D
    $\left(\frac{-4}{5}, \frac{7}{5}\right)$
Answer
Correct option: A.
$\left(\frac{4}{5}, \frac{7}{5}\right)$
(a) : Given equation of circle is
$x^2+y^2-4 x-6 y-12=0$
i.e., $(x-2)^2+(y-3)^2=5^2$.....(i)
So the given circle has centre $B(2,3)$ and radius 5 units.
Let $A(h, k)$ be the centre of circle having radius 3 units and touches the circle (i) at $P(-1,-1)$
In the given diagram, we have $B P=5$ and $A P=3$
$\therefore \quad A B=B P-A P=5-3=2$
$\therefore \quad A(h, k)$ divides $B P$ in the ratio $3: 2$ internally
$
\therefore h=\frac{3 \times 2+2 \times(-1)}{3+2}=\frac{4}{5} ; k=\frac{3 \times 3+2 \times(-1)}{3+2}=\frac{7}{5}
$
$\therefore \quad$ Required centre of circle is $\left(\frac{4}{5}, \frac{7}{5}\right)$.
https://pg-data.sgp1.digitaloceanspaces.com/chapter_wise/16527/D5.png" alt="Image" width="110" height="">
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MCQ 132 Marks
If the line $x-2 y=m(m \in Z )$ intersects the circle $x^2+y^2=2 x+4 y$ at two distinct points, then the number of possible values of $m$ are
  • A
    8
  • ✓
    9
  • C
    10
  • D
    11
Answer
Correct option: B.
9
(b) : Equation of circle is $x^2+y^2=2 x+4 y$, can be written as $(x-1)^2+(y-2)^2=5$
$\therefore \quad$ Centre of circle is $(1,2)$ and radius is $\sqrt{5}$.
Distance of $(1,2)$ from line $x-2 y-m=0$
$=\left|\frac{1-4-m}{\sqrt{1^2+2^2}}\right|<\sqrt{5} \quad$ (as line intersect the circle)
$
\Rightarrow|-3-m|<5 \Rightarrow|-(m+3)|<5 \Rightarrow m+3<5
$
$
m=-7,-6,-5,-4,-3,-2,-1,0,1 \text {. }
$
$\therefore \quad$ Number of possible values of $m$ is 9 .
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MCQ 142 Marks
The equation of the circle whose centre lies on the line $x-4 y=1$ and which passes through the points $(3,7)$ and $(5,5)$ is
  • ✓
    $x^2+y^2+6 x+2 y-90=0$
  • B
    $x^2+y^2+6 x-2 y+90=0$
  • C
    $x^2+y^2+6 x+2 y+90=0$
  • D
    $x^2+y^2-6 x+2 y-90=0$
Answer
Correct option: A.
$x^2+y^2+6 x+2 y-90=0$
(a) : Consider the line first, we have $x-4 y=1$
$\Rightarrow x=1+4 y$
$\therefore \quad$ Any point on the line is of the form $(4 a+1, a)$, which included the centre of circle also. This point is equidistant from any point on the circumference of the circle.
$
\begin{array}{r}
\therefore \quad(4 a+1-3)^2+(a-7)^2=(4 a+1-5)^2+(a-5)^2 \\
\Rightarrow \quad 16 a^2-16 a+4+a^2-14 a+49=16 a^2-32 a+16 \\
+a^2-10 a+25
\end{array}
$
$
\begin{aligned}
& \Rightarrow \quad-12 a=12 \Rightarrow a=-1 \\
& \therefore \text { Centre of circle }=(-4+1,-1)=(-3,-1)
\end{aligned}
$ Now, radius $(r)=\sqrt{(-3-3)^2+(-1-7)^2}=\sqrt{100}=10$
$\therefore \quad$ Equation of circle is $(x+3)^2+(y+1)^2=10^2$
i.e., $x^2+y^2+6 x+2 y-90=0$
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MCQ 152 Marks
The parametric equation of the circle $x^2+y^2-6 x-$ $2 y+9=0$ are
  • ✓
    $x=3+\cos \theta, y=1+\sin \theta$
  • B
    $x=1+\cos \theta, y=3+\sin \theta$
  • C
    $x=\cos \theta, y=\sin \theta$
  • D
    $x=3+\sin \theta, y=1+\cos \theta$
Answer
Correct option: A.
$x=3+\cos \theta, y=1+\sin \theta$
(a) : If $\left(x_1, y_1\right)$ is centre of circle and $r$ is radius then parametric equation of circle can be written as
$
x=x_1+r \cos \theta, \quad y=y_1+r \sin \theta
$
Given equation of circle is
$
x^2+y^2-6 x-2 y+9=0 \Rightarrow(x-3)^2+(y-1)^2=1
$
$\therefore \quad$ Centre of circle is $(3,1)$ and radius of circle is 1 .
Equation in parametric form is $x=3+\cos \theta, \quad y=1+\sin \theta$
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MCQ 162 Marks
The equation of tangents to the circle $x^2+y^2=4$ which are parallel to $x+2 y+3=0$ are
  • A
    $x-2 y= \pm 2 \sqrt{5}$
  • B
    $x-2 y= \pm 2$
  • C
    $x+2 y= \pm 2 \sqrt{3}$
  • ✓
    $x+2 y= \pm 2 \sqrt{5}$
Answer
Correct option: D.
$x+2 y= \pm 2 \sqrt{5}$
(d) : Given, equation of the circle is $x^2+y^2=4$ and equation of the line is $x+2 y+3=0$
Slope of the line $=-1 / 2$
Since, the required tangents are parallel to the given line.
$\therefore \quad$ Slope of tangents $(m)=-\frac{1}{2}$
We know that the equation of the tangents to the circle $x^2+y^2=a^2$ with slope $m$ are $y=m x \pm \sqrt{a^2\left(1+m^2\right)}$
$\therefore \quad$ The required equation of tangents are
$
\begin{aligned}
y & =-\frac{1}{2} x \pm \sqrt{(2)^2\left(1+\left(\frac{-1}{2}\right)^2\right)} \\
\Rightarrow y & =-\frac{1}{2} x \pm \sqrt{5} \Rightarrow 2 y+x= \pm 2 \sqrt{5}
\end{aligned}
$
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MCQ 172 Marks
Find the equation of a circle of radius 5 units whose centre lies on $x$-axis and passes through the point $(2,4)$.
  • A
    $x^2+y^2-12 x-11=0$
  • B
    $x^2+y^2-4 x-21=0$
  • ✓
    $x^2+y^2+2 x-24=0$
  • D
    $x^2+y^2+12 x-11=0$
Answer
Correct option: C.
$x^2+y^2+2 x-24=0$
(c) : Let the coordinates of the centre of the required circle be $C(a, 0)$. Since it passes through $P(2,4)$.
$\therefore C P=$ radius $\Rightarrow C P=5$
$\Rightarrow \sqrt{(a-2)^2+(0-4)^2}=5$
$\Rightarrow(a-2)^2+16=25 \Rightarrow a-2= \pm 3 \Rightarrow a=5$ or $a=-1$
Thus, the coordinates of the centre are $(5,0)$ or $(-1,0)$. Hence, the equations of the required circle are
$
(x-5)^2+(y-0)^2=5^2 \text { and }(x+1)^2+(y-0)^2=5^2
$
$\Rightarrow x^2+y^2-10 x=0$ and $x^2+y^2+2 x-24=0$
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MCQ 182 Marks
The intercept on the line $y=x$ by the circle $x^2+y^2-2 x=0$ is $A B$. The equation of the circle with $A B$ as a diameter is
  • A
    $x^2+y^2+x+y=0$
  • ✓
    $x^2+y^2-x-y=0$
  • C
    $x^2+y^2-3 x+y=0$
  • D
    $x^2+y^2+3 x-y=0$
Answer
Correct option: B.
$x^2+y^2-x-y=0$
(b) : Points of intersection of the line $y=x$ and circle $x^2+y^2-2 x=0$ are given by $x^2+x^2-2 x=0 \Rightarrow 2 x^2-2 x=0$ $\Rightarrow 2 x(x-1)=0 \Rightarrow x=0$ or $x=1$
When $x=0, y=0$, When $x=1, y=1$
$\therefore \quad$ Points $A$ and $B$ are $(0,0)$ and $(1,1)$
Now, the equation of a circle whose diameter is $A B$ is given by $(x-0)(x-1)+(y-0)(y-1)=0$
$
\Rightarrow x^2-x+y^2-y=0 \Rightarrow x^2+y^2-x-y=0
$
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MCQ 192 Marks
The sides of a rectangle are given by $x= \pm a$ and $y= \pm b$. The equation of the circle passing through the vertices of the rectangle is
  • A
    $x^2+y^2=a^2$
  • ✓
    $x^2+y^2=a^2+b^2$
  • C
    $x^2+y^2=a^2-b^2$
  • D
    $(x-a)^2+(y-b)^2=a^2+b^2$
Answer
Correct option: B.
$x^2+y^2=a^2+b^2$
(b) : Centre of the required circle $=(0,0)$.
Radius $(r)=\sqrt{(a-0)^2+(b-0)^2}$
$=\sqrt{a^2+b^2}$
$\therefore$ Equation of circle is $(x-0)^2+(y-0)^2=\left(\sqrt{a^2+b^2}\right)^2 \Rightarrow x^2+y^2=a^2+b^2$
Image
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MCQ 202 Marks
The angle between a pair of tangents drawn from a point P to the circle $x^2+y^2+4 x-6 y+9 \sin ^2 \alpha+13 \cos ^2 \alpha=0$ is $2 \alpha$.The equation of the locus of the point P is
  • A
    $x^2+y^2+4 x-6 y+4=0$
  • B
    $x^2+y^2+4 x-6 y-9=0$
  • C
    $x^2+y^2+4 x-6 y-4=0$
  • D
    $x^2+y^2+4 x-6 y+9=0$
Answer
(D)
The centre of the circle
$x^2+y^2+4 x-6 y+9 \sin ^2 \alpha+13 \cos ^2 \alpha=0$
is $C (-2,3)$ and its radius is
$\begin{array}{l}\sqrt{2^2+(-3)^2-9 \sin ^2 \alpha-13 \cos ^2 \alpha} \\=\sqrt{4+9-9 \sin ^2 \alpha-13 \cos ^2 \alpha}=2 \sin \alpha\end{array}$
Let $P ( h , k )$ be any point on the locus. Then $\angle APC =\alpha$. Also $\angle PAC =\pi / 2$ i.e., triangle APC is a right angle triangle.
Image
$\therefore \sin \alpha=\frac{AC}{PC}=\frac{2 \sin \alpha}{\sqrt{(h+2)^2+(k-3)^2}}$
$ \Rightarrow \sqrt{(h+2)^2+(k-3)^2}=2 $
$\Rightarrow(h+2)^2+(k-3)^2=4$
$\Rightarrow h^2+k^2+4 h-6 k+9=0$
∴ The required equation of the locus is
$x^2+y^2+4 x-6 y+9=0$
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MCQ 212 Marks
Two tangents PQ and PR drawn to the circle $x^2+y^2-2 x-4 y-20=0$ from point P(16, 7). If the centre of the circle is C, then the area of quadrilateral PQCR will be
  • A
    150 sq. units
  • B
    50 sq. units
  • ✓
    75 sq. units
  • D
    70 sq. units.
Answer
Correct option: C.
75 sq. units
(C)
Required area $=\frac{1}{2}(5 \times 15) \times 2=75$
Image
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MCQ 222 Marks
If OA and OB are the tangents from the origin to the circle $x^2+y^2+2 g x+2 f y+c=0$ and C is the centre of the circle, the area of the quadrilateral OACB is
  • A
    $\frac{1}{2} \sqrt{c\left(g^2+f^2-c\right)}$
  • B
    $\sqrt{c\left(g^2+f^2-c\right)}$
  • C
    $c \sqrt{g^2+f^2-c}$
  • D
    $\frac{\sqrt{g^2+f^2-c}}{c}$
Answer
(B)
Let $S \equiv x^2+y^2+2 gx +2 fy + c =0$
Area of quadrilateral $=2 \ [area \ of \ \triangle OAC ]$
$=2 \cdot \frac{1}{2} OA \cdot AC=\sqrt{S_1} \cdot \sqrt{g^2+f^2-c}$
Image
Point O is $(0,0)$
$\Rightarrow S _1= c$
$\therefore $ Area $=\sqrt{ c } \cdot \sqrt{ g ^2+ f ^2- c }$
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MCQ 232 Marks
Let the tangents drawn from the origin to the circle, $x^2+y^2-8 x-4 y+16=0$ touch it at the points A and B. The $(A B)^2$ is equal to
  • A
    $\frac{56}{5}$
  • ✓
    $\frac{64}{5}$
  • C
    $\frac{32}{5}$
  • D
    $\frac{52}{5}$
Answer
Correct option: B.
$\frac{64}{5}$
(B)
Image
$\begin{array}{l}\text { Radius }= R =\sqrt{16+4-16}=2 \\ L= OA =\sqrt{ S _1}=\sqrt{16}=4 \\ \text { Length of } AB =\frac{2 LR }{\sqrt{ L ^2+ R ^2}} \\ \quad=\frac{2 \times 4 \times 2}{\sqrt{4^2+2^2}}=\frac{16}{\sqrt{20}}=\frac{8}{\sqrt{5}}\end{array}$
$\therefore AB ^2=\frac{64}{5}$
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MCQ 242 Marks
The equations of the tangents draws from the point (0, 1) to the circle $x^2+y^2-2 x+4=0$ are
  • ✓
    $2 x-y+1=0, x+2 y-2=0$
  • B
    $2 x-y+1=0, x+2 y+2=0$
  • C
    $2 x-y-1=0, x+2 y-2=0$
  • D
    $2 x-y-1=0, x+2 y+2=0$
Answer
Correct option: A.
$2 x-y+1=0, x+2 y-2=0$
(A)
Required equations are given by $SS _1= T ^2$
$ \Rightarrow\left(x^2+y^2-2 x+4 y\right)(1+4)=\{y-1(x)+2(y+1)\}^2$
$\Rightarrow 2 x^2-2 y^2-3 x+4 y+3 x y-2=0$
$\Rightarrow(2 x-y+1)(x+2 y-2)=0$
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MCQ 252 Marks
The equations of the tangets drawn the origin to the circle $x^2+y^2-2 r x-2 h y+h^2=0$ are
  • A
    $x=0, y=0$
  • ✓
    $\left( h ^2- r ^2\right) x-2 rh y=0, x=0$
  • C
    $y=0, x=4$
  • D
    $h ^2- r ^2 x+2 rh y=0, x=0$
Answer
Correct option: B.
$\left( h ^2- r ^2\right) x-2 rh y=0, x=0$
(B)
Required equations are given by $SS _1= T ^2$
$\Rightarrow h^2\left(x^2+y^2-2 r x-2 h y+h^2\right)=\left(r x+h y-h^2\right)^2$
$\Rightarrow\left(h^2-r^2\right) x^2-2 rh x y=0$
$\Rightarrow x\left\{\left(h^2-r^2\right) x-2 rh y\right\}=0$
$\Rightarrow x=0,\left(h^2-r^2\right) x-2 rh y=0$
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MCQ 262 Marks
If tangents are drawn to the circle $x^2+y^2=12$ at the points where it intersects the circle $x^2+y^2-5 x+3 y-2=0$, then the coordinates of the point of intersection of those tangents are
  • A
    $\left(-6, \frac{18}{5}\right)$
  • B
    $\left(6, \frac{18}{5}\right)$
  • C
    $\left(-6, \frac{-18}{5}\right)$
  • D
    $\left(6, \frac{-18}{5}\right)$
Answer
(D)
Let $( h , k )$ be the point of intersection.
Then the chord of the tangent is the common chord of the circles $x^2+y^2=12$ and
Image
The equation of the common chord is
If $S_1$ and $S_2$ are two circles, then equation of their common chord is $S _1- S _2=0$.
i.e., $2\left(g_1- g _2\right) x+2\left( f _1- f _2\right) y+\left( c _1- c _2\right)=0$
$5 x-3 y-10=0 \quad\ldots(i)$
The equation of common chord of contact is
$h x+k y-12=0 \quad\ldots(ii)$
Equations (i) and (ii) represents the same line.
$\begin{array}{l}\therefore \frac{h}{5}=\frac{k}{-3}=\frac{-12}{-10} \\\Rightarrow h=6, k=\frac{-18}{5}\end{array}$
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MCQ 272 Marks
The equations of tangents to the circle $x^2+y^2-22 x-4 y+25=0$ which are perpendicular to the line $5 x+12 y+8=0$ are
  • ✓
    $12 x-5 y+8=0,12 x-5 y=252$
  • B
    $12 x-5 y=0,12 x-5 y=252$
  • C
    $12 x-5 y-8=0,12 x-5 y+252=0$
  • D
    $12 x-5 y+25=0,12 x-5 y+16=0$
Answer
Correct option: A.
$12 x-5 y+8=0,12 x-5 y=252$
(A)
Equation of line perpendicular to
$5 x+12 y+8=0 \text { is } 12 x-5 y+k=0 .$
Now it is a tangent to the circle, if
Radius of circle $=$ Distance of line from centre of circle$\sqrt{121+4-25}=\left|\frac{12(11)-5(2)+k}{\sqrt{144+25}}\right|$
$\Rightarrow k =8$ or -252 .
Hence, equations of tangents are
$12 x-5 y+8=0,12 x-5 y=252$
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MCQ 282 Marks
The equation of the tangent to the circle $x^2+y^2-2 x-4 y-4=0$ which is perpendicular to $3 x-4 y-1=0$, is
  • A
    $4 x+3 y-15=0$
  • B
    $4 x+3 y+25=0$
  • C
    $4 x-3 y+15=0$
  • ✓
    $4 x+3 y-25=0$
Answer
Correct option: D.
$4 x+3 y-25=0$
(D)
Image
Tangent is of form $4 x+3 y+ c =0$. From condition of tangency to the circle, we get $c =-25$. Hence, equation is $4 x+3 y-25=0$.
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MCQ 292 Marks
lf $a>2 b>0$, then the positive value of m for which $y=m x-b \sqrt{1+m^3}$ is a common tangent to $x^2+y^2= b ^2$ and $(x- a )^2+y^2= b ^2$, is
  • ✓
    $\frac{2 b}{\sqrt{a^2-4 b^2}}$
  • B
    $\frac{\sqrt{a^3-4 b^2}}{2 b}$
  • C
    $\frac{2 b}{a-2 b}$
  • D
    $\frac{b}{a-2 b}$
Answer
Correct option: A.
$\frac{2 b}{\sqrt{a^2-4 b^2}}$
(A)
Any tangent to $x^2+y^2= b ^2$ is $y= m x- b \sqrt{1+ m ^2}$
It touches $(x- a )^2+y^2= b ^2$
if $\frac{m a-b \sqrt{1+m^2}}{\sqrt{m^2+1}}=b$ or $m a=2 b \sqrt{1+m^3}$
or $m^2 a^2=4 b^2\left(1+m^2\right)$,
$\therefore$ $m= \pm \frac{2 b}{\sqrt{a^2-4 b^2}}$
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MCQ 302 Marks
Consider the circle $x^2+y^2-6 x+4 y=12$. The equation of a tangent to this circle that is parallel to the line $4 x+3 y+5=0$ is
  • A
    $4 x+3 y+10=0$
  • B
    $4 x+3 y-9=0$
  • C
    $4 x+3 y+9=0$
  • ✓
    $4 x+3 y-31=0$
Answer
Correct option: D.
$4 x+3 y-31=0$
(D)
Let the equation of tangent be
$\begin{array}{l}4 x+3 y+k=0 ...(i) \\S=x^2+y^2-6 x+4 y-12=0\end{array}$
The centre and radius of S are $(3,-2)$ and 5 units
Distance of (i) from centre of $S =$ radius
$\begin{array}{l}\Rightarrow\left|\frac{12-6+k}{5}\right|=5 \\\Rightarrow|6+k|=25 \\\Rightarrow 6+k= \pm 25 \\ \Rightarrow k=19 \text { or }-31\end{array}$
∴ Equation of tangent is $4 x+3 y+19=0$ or$4 x+3 y-31=0$
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MCQ 312 Marks
The equations of the tangents to the circle $x^2+y^2=a^2$ parallel to the line $\sqrt{3} x+y+3=0$ are
  • ✓
    $\sqrt{3} x+y \pm 2 a=0$
  • B
    $\sqrt{3} x+y \pm a =0$
  • C
    $\sqrt{3} x+y \pm 4 a=0$
  • D
    $\sqrt{3} y-x+3=0$
Answer
Correct option: A.
$\sqrt{3} x+y \pm 2 a=0$
(A)
Equation of line parallel to the
$\sqrt{3} x+y+3=0 \text { is } \sqrt{3} x+y+k=0$
But it is a tangent to the circle
$x^2+y^2= a ^2$, then
$\begin{array}{l}\left|\frac{k}{\sqrt{1+3}}\right|=a \\\Rightarrow k= \pm 2 a\end{array}$
Hence, the required equation is
$\sqrt{3} x+y \pm 2 a=0$
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MCQ 322 Marks
If the equation of the tangent to the circle $x^2+y^2-2 x+6 y-6=0$ parallel to $3 x-4 y+7=0$ is $3 x-4 y+k=0$, then the values of k are
  • ✓
    5, -35
  • B
    -5, 35
  • C
    7, -32
  • D
    -7, 32
Answer
Correct option: A.
5, -35
(A)
Image
$3 x-4 y+ k =0$
Equation of circle is,
$x^2+y^2-2 x+6 y-6=0$
centre $(1,-3)$
Radius of circle $=4$
And centre of circle $=(1,-3)$
Equation of tangent is $3 x-4 y+ k =0$
$\therefore \frac{3 \times 1-4 \times(-3)+k}{\sqrt{(3)^2+(-4)^2}}= \pm 4$
Hence, $k =5,-35$
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MCQ 332 Marks
The equations of the tangents to circle $5 x^2+5 y^2=1$, parallel to line 3x + 4y = 1 are
  • A
    $3 x+4 y= \pm 2 \sqrt{5}$
  • B
    $6 x+8 y= \pm \sqrt{8}$
  • ✓
    $3 x+4 y= \pm \sqrt{5}$
  • D
    $6 x+8 y= \pm \sqrt{5}$
Answer
Correct option: C.
$3 x+4 y= \pm \sqrt{5}$
(C)
Equations of tangents are
$y=\frac{-3}{4} x \pm \frac{1}{\sqrt{5}} \sqrt{1+\left(\frac{-3}{4}\right)^2}$
[∵ equation of tangent is $y= m x \pm a \sqrt{1+ m ^2}$ ]
$\begin{array}{ll}\therefore & y=\frac{-3}{4} x \pm \frac{1}{\sqrt{5}} \sqrt{\frac{16+9}{16}} \\ \therefore & 4 y=-3 x \pm \sqrt{5} \\ & \Rightarrow 3 x+4 y= \pm \sqrt{5}\end{array}$
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MCQ 342 Marks
The equation of the tangents to the circle $x^2+y^2-20 x+12 y+11=0$ having slope -2 are
  • A
    $y=-2 x+11$ and $y=-2 x+39$
  • B
    $y=-2 x-15$ and $y=-2 x+39$
  • C
    $y=-2 x-11$ and $y=-2 x+35$
  • ✓
    $y=-2 x-11$ and $y=-2 x+39$
Answer
Correct option: D.
$y=-2 x-11$ and $y=-2 x+39$
(D)
Image
Equation of tangent $y= m x+ c$, where $m =-2$ is $y=-2 x+ c$
$\therefore 2 x+y-c=0$
Now, $x^2+y^2-20 x+12 y+11=0$
$\therefore 2 g=-20,2 f =12$
$\Rightarrow g =-10, f =6, c =11$
$\therefore$ Centre $(- g ,- f )=(10,-6)$
$\text { radius }=\sqrt{g^2+f^2-c}=\sqrt{125}=5 \sqrt{5}$
Since distance of tangent from the centre is equal to radius,
$\begin{array}{l} 5 \sqrt{5}=\left|\frac{2(10)-6-c}{\sqrt{4+1}}\right| \\ \Rightarrow 5 \sqrt{5} \times \sqrt{5}=|14-c| \\ \Rightarrow c=-11 \text { or } 39\end{array}$
∴ Equations of tangents are
$y=-2 x-11$ and $y=-2 x+39$
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MCQ 352 Marks
Tangents are drawn from the origin to a circle with centre at (2, -1). If the equation of one of the tangents is 3x + y = 0, the equation of the other tangent is
  • A
    3x - y = 0
  • B
    x + 3y = 0
  • ✓
    x - 3y = 0
  • D
    x + 2y = 0
Answer
Correct option: C.
x - 3y = 0
(C)
Let the equation of other tangent from the origin be $y=m x$, then length of the perpendiculars from the centre $(2,-1)$ on the two tangents is same.
$\begin{array}{l}\left|\frac{2 m+1}{\sqrt{1+m^2}}\right|=\left|\frac{6-1}{\sqrt{9+1}}\right|=\frac{5}{\sqrt{10}} \\\Rightarrow m=-3 \text { or } \frac{1}{3}\end{array}$
∴ Slope of other tangent is $\frac{1}{3}$ and its equation is
$\begin{array}{l}y=\left(\frac{1}{3}\right) x \\\Rightarrow x-3 y=0\end{array}$
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MCQ 362 Marks
If the squares of the lengths tangents a point circles $x^2+y^2=a^2, x^2+y^2=b^2$ and $x^2+y^2=c^2$ are in A.P. then
  • A
    $a , b , c$ are in G.P.
  • B
    $a , b , c$ are in A.P.
  • ✓
    $a^2, b^2, c^2$ are in A.P.
  • D
    $a^2, b^2, c^2$ are in G.P.
Answer
Correct option: C.
$a^2, b^2, c^2$ are in A.P.
(C)
Let $P \left(x_1, y_1\right)$ be a point. Let $l_1^2, l_2^2, l_3^2$ be the squares of lengths of the tangents from the point P $\left(x_1, y_1\right)$.
$\begin{aligned}\therefore & l_1^2=x_1^2+y_1^2-a^2 \\& l_2^2=x_1^2+y_1^2-b^2 \\&l_3^2=x_1^2+y_1^2-c^2\end{aligned}$
Assume that $x_1^2+y_1^2= k$
$\therefore l_1^2=k-a^2 ; l_2^2=k-b^2 ; l_3^2=k-c^2$
or we can say that
$a^2=k-l_1^2 ; b^2=k-l_2^2 ; c^2=k-l_3^2$
Since $l_1^2, l_2^2$ and $l_3^2$ are in A. P., we have
$a ^2, b^2$ and $c ^2$ are in A.P.
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MCQ 372 Marks
The co-ordinates of the point from where the tangents are drawn to the circles, $x^2+y^2=1$, $x^2+y^2+8 x+15=0$ and $x^2+y^2+10 y+24=0$ are of same length ,are
  • A
    $\left(2, \frac{5}{2}\right)$
  • ✓
    $\left(-2, \frac{5}{2}\right)$
  • C
    $\left(-2, \frac{5}{2}\right)$
  • D
    $\left(2,-\frac{5}{2}\right)$
Answer
Correct option: B.
$\left(-2, \frac{5}{2}\right)$
(B)
Length of tangents is same i.e.,$\sqrt{S_1}=\sqrt{S_2}=\sqrt{S_3}$
We get the point from where tangent is drawn, by solving the 3 equations for $x$ and $y$.
i.e.,,$x^2+y^2=1$,
$x^2+y^2+8 x+15=0$ and
$x^2+y^2+10 y+24=0$
or $8 x+16=0$ and $10 y+25=0$
$\Rightarrow x=-2$ and $y=-\frac{5}{2}$
Hence, the point is $\left(-2,-\frac{5}{2}\right)$.
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MCQ 382 Marks
If the lengths of the tangents drawn from P to the circles $x^2+y^2-2 x+4 y-20=0$ and $x^2+y^2-2 x-8 y+1=0$ are in thr ratio 2 : 1 , then the locus of P is
  • A
    $x^2+y^2+2 x+12 y+8=0$
  • B
    $x^2+y^2-2 x+12 y+8=0$
  • C
    $x^2+y^2+2 x-12 y+8=0$
  • ✓
    $x^2+y^2-2 x-12 y+8=0$
Answer
Correct option: D.
$x^2+y^2-2 x-12 y+8=0$
(D)
(Let the point be $P\left(x_1, y_1\right)$
According to the given condition,
$\begin{array}{l}\frac{\sqrt{x_1^2+y_1^2-2 x_1+4 y_1-20}}{\sqrt{x_1^2+y_1^2-2 x_1-8 y_1+1}}-\frac{2}{1} \\\Rightarrow \frac{x_1^2+y_1^2-2 x_1+4 y_1-20}{x_1^2+y_1^2-2 x_1-8 y_1+1}=\frac{4}{1} \\\Rightarrow x_1^2+y_1^2-2 x_1+4 y_1-20 =4 x_1^2+4 y_1^2-8 x_1-32 y_1+4 \\\Rightarrow 3 x_1^2+3 y_1^2-6 x_1-36 y_1+24=0 \\\Rightarrow x_1^2+y_1^2-2 x_1-12 y_1+8=0\end{array}$
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MCQ 392 Marks
If P is a point such that the ratio of the squares of the lengths of the tangents from P to the circles $x^2+y^2+2 x-4 y-20=0$ and $x^2+y^2-4 x+2 y-44=0$ is $ 2: 3 $, then the locus of Pis a circle with centre
  • A
    (7 , -8)
  • ✓
    (-7 , 8)
  • C
    (7 , 8)
  • D
    (-7 , -8)
Answer
Correct option: B.
(-7 , 8)
(B)
$\frac{x^2+y^2+2 x-4 y-20}{x^2+y^2-4 x+2 y-44}=\frac{2}{3}$
$\Rightarrow x^2+y^2+14 x-16 y+28=0$
$\therefore $ Centre $=(-7,8)$
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MCQ 402 Marks
If the ratio of the lengths of tangents drawn from the point (f, g) to the given circle $x^2+y^2=6$ and $x^2+y^2+3 x+3 y=0$ be $ 2: 1 $, then
  • A
    $f^2+g^2+2 g+2 f+2=0$
  • B
    $f^2+g^2+4 g+4 f+4=0$
  • ✓
    $f^2+g^2+4 g+4 f+2=0$
  • D
    $f^2+g^2+4 g+2 f+4=0$
Answer
Correct option: C.
$f^2+g^2+4 g+4 f+2=0$
(C)
According to the given condition,
$\frac{f^2+g^2-6}{f^2+g^2+3 f+3 g}=\frac{4}{1}$
$\Rightarrow f^2+g^2+4 f+4 g+2=0$
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MCQ 412 Marks
Given the circles $x^2+y^2-4 x-5=0$ and $x^2+y^2+6 x-2 y+6=0$ let p be a point $(\alpha, \beta)$, such that the tangents from P to both the circles are equal, then
  • A
    $2 \alpha+10 \beta+11=0$
  • B
    $2 \alpha-10 \beta+11=0$
  • ✓
    $10 \alpha-2 \beta+11=0$
  • D
    $10 \alpha+2 \beta+11=0$
Answer
Correct option: C.
$10 \alpha-2 \beta+11=0$
(C)
According to the given condition,
$\alpha^2+\beta^2-4 \alpha-5=\alpha^2+\beta^2+6 \alpha-2 \beta+6$
$\therefore 10 \alpha-2 \beta+11=0$
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MCQ 422 Marks
The length of the tangents drawn from any point on the circle $x^2+y^2+2 g x+2 f y+ C _1=0$ to the circle $x^2+y^2+2 g x+2 f y+C_2=0$ is
  • A
    $\sqrt{C_1^2+C_2^2}$
  • ✓
    $\sqrt{C_2-C_1}$
  • C
    $C_1+C_2$
  • D
    $C _1= C _2$
Answer
Correct option: B.
$\sqrt{C_2-C_1}$
(B)
Let $\left(x_1, y_1\right)$ be any point on the circle
$x^2+y^2+2 g x+2 f y+ C _1=0$
$\therefore x_1^2+y_1^2+2 g x_1+2 f y_1+ C _1=0$
i.e. $x_1^2+y_1^2+2 g x_1+2 f y_1=- C _1$
Length of the tangent from $\left(x_1, y_1\right)$ to the circle
$x^2+y^2+2 g x+2 f y+ C _2=0$ is
$\begin{aligned}\sqrt{x_1^2+y_1^2+2 g x_1+2 f y_1+C_2} & =\sqrt{-C_1+C_2} \\& =\sqrt{C_2-C_1}\end{aligned}$
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MCQ 432 Marks
If the tangent at the point P on the circle $x^2+y^2+6 x+6 y=2$ meets the straight line $5 x-2 y+6=0$ at a point Q on the Y-axis, then the length of PQ is
  • A
    $2 \sqrt{5}$
  • B
    $3 \sqrt{5}$
  • C
    $4$
  • ✓
    $5$
Answer
Correct option: D.
$5$
(D)
Tangent of the given circle meets the line $5 x-2 y+6=0$ at a point $Q (0,3)$ on the Y -axis.
$\therefore$ Length of tangent $=\sqrt{(0)^2+(3)^2+6(0)+6(3)-2}$
$\begin{aligned}PO & =\sqrt{9+18-2} \\& =\sqrt{25}=5\end{aligned}$
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MCQ 442 Marks
If circles $x^2+y^2-4 x-6 y+9=0$ and $x^2+y^2+2 x+2 y-7=0$ touch each other, then their point of contact is
  • A
    $\left(\frac{4}{5}, \frac{-7}{5}\right)$
  • B
    $\left(\frac{-4}{5}, \frac{7}{5}\right)$
  • C
    $\left(\frac{-4}{5}, \frac{-7}{5}\right)$
  • ✓
    $\left(\frac{4}{5}, \frac{7}{5}\right)$
Answer
Correct option: D.
$\left(\frac{4}{5}, \frac{7}{5}\right)$
(D)
Let $S _1: x^2+y^2-4 x-6 y+9=0$ and $S _2: x^2+y^2+2 x+2 y-7=0$
$\therefore$ the equation of the common tangent to both the circles is $S _1- S _2=0$
$\Rightarrow 6 x+8 y-16=0$
$\Rightarrow 3 x+4 y=8$ is the common equation of the tangent from the given choices,
it is clear that the point $\left(\frac{4}{5}, \frac{7}{5}\right)$ lies on the tangent.
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MCQ 452 Marks
If the circles given by $S \equiv x^2+y^2-14 x+6 y+33=0$ and $S^{\prime} \equiv x^2+y^2-a^2=0(a \in N)$ have 4 common tangents, then the pissible number of circles $S ^{\prime}=0$ is
  • A
    $1$
  • ✓
    $2$
  • C
    $0$
  • D
    infinite
Answer
Correct option: B.
$2$
(B)
The centres and radii of the two circles are
$C _1(7,-3), C _2(0,0), r _1=5, r _2= a$
For 4 common tangents,
$r_1+r_2<\left|C_1 C_2\right|$
$C_1 C_2=\sqrt{7^2+(-3)^2}=\sqrt{58} \approx 7.6$
$r _1+ r _2<\left| C _1 C _2\right|$ for $a =1,2$
$\therefore$ Number of possible circles $=2$
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MCQ 462 Marks
The number of common tangents to the circles $x^2+y^2-4 x-6 y-12=0$ and $x^2+y^2+6 x+18 y+26=0$, is
  • A
    $1$
  • B
    $2$
  • ✓
    $3$
  • D
    $4$
Answer
Correct option: C.
$3$
(C)
$x^2+y^2-4 x-6 y-12=0$
$\begin{array}{l} C_1=(2,3), r_1=\sqrt{4+9+12}=5 \\ x^2+y^2+6 x+18 y+26=0 \\ C_2=(-3,-9), r_2=\sqrt{9+81-26}=8 \\ C_1 C_2=\sqrt{(-3-2)^2+(-9-3)^2}=13 \\ C_1 C_2=r_1+r_2 \\ \Rightarrow \text { The given circles touch each other externally. } \\ \Rightarrow \text { Number of common tangents is } 3\end{array}$
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MCQ 472 Marks
The number of common tangents to circles $x^2+y^2+2 x+8 y-23=0$ and
$x^2+y^2-4 x-10 y+9=0$ is
  • A
    $1$
  • B
    $3$
  • ✓
    $2$
  • D
    None of these
Answer
Correct option: C.
$2$
(C)
$x^2+y^2+2 x+8 y-23=0$
$\therefore \quad C _1(-1,-4), r _1=2 \sqrt{10}$
Again $x^2+y^2-4 x-10 y+9=0$
$\therefore C _2(2,5), r _2=2 \sqrt{5}$
Now $C _1 C _2$ = distance between centres.
$\therefore C _1 C _2=\sqrt{9+81}=3 \sqrt{10}=9.486$ and
$r_1+r_2=2(\sqrt{10}+\sqrt{5})=10.6$
$r_1-r_2=2 \sqrt{5}(\sqrt{2}-1)$
$=2 \times 2.2 \times 0.4$
$=4.4 \times 0.4$
$=1.76$
$C _1 C _2=2 \sqrt{10}> r _1- r _2$
$r_1-r_2 < C_1 C_2 < r_1+r_2$
⇒ Two tangents can be drawn.
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MCQ 482 Marks
If a circle, whose centre is (-1, 1) touches the straight line x + 2y + 12 = 0 then the co-ordinates of the point of contact are
  • A
    $\left(\frac{-7}{2},-4\right)$
  • ✓
    $\left(\frac{-18}{5}, \frac{-21}{5}\right)$
  • C
    (2, -7)
  • D
    (-2, -5)
Answer
Correct option: B.
$\left(\frac{-18}{5}, \frac{-21}{5}\right)$
(B)
Image

Let point of contact be $P \left(x_1, y_1\right)$.
This point lies on line $x_1+2 y_1=-12$ ...(i)
Gradient of OP $= m _1=\frac{y_1-1}{x_1+1}$
Gradient of $x+2 y+12= m _2=-\frac{1}{2}$
The two lines are perpendicular.
$\therefore \left(\frac{y_1-1}{x_1+1}\right)\left(\frac{-1}{2}\right)=-1$
$\Rightarrow 2 x_1-y_1=-3$ ...(ii)
On solving equations (i) and (ii), we get
$\left(x_1, y_1\right)=\left(\frac{-18}{5}, \frac{-21}{5}\right)$
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MCQ 492 Marks
The point of contact of the tangent to the circle $x^2+y^2=5$ at the point (1,-2) which touches the circle ,$x^2+y^2-8 x+6 y+20=0$ is
  • A
    (2, -1)
  • ✓
    (3 ,-1)
  • C
    (4, -1)
  • D
    (5, -1)
Answer
Correct option: B.
(3 ,-1)
(B)
Equation of the tangent at $(1,-2)$ to the circle $x^2+y^2=5$ is $x-2 y=5$
Here, only point $(3,-1)$ lies on the tangent.
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MCQ 502 Marks
The line $x \cos \alpha+y \sin \alpha= p$ will be a tangent to the circle $x^2+y^2-2 a x \cos \alpha-2 a y \sin \alpha=0$, if $p =$
  • A
    $0$ or a
  • B
    $0$
  • C
    2a
  • ✓
    $0$ or 2a
Answer
Correct option: D.
$0$ or 2a
(D)
$x \cos \alpha+y \sin \alpha- p =0$ is a tangent, if perpendicular from centre on it is equal to radius of the circle. Here centre is
( $a \cos \alpha, a \sin \alpha$ ) and radius is a.
$\therefore$ $\left|\frac{a \cos ^2 \alpha+a \sin ^2 \alpha-p}{\sqrt{1}}\right|=a$
i.e., $|a-p|=a \Rightarrow p=0$ or $p=2 a$
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MCQ 512 Marks
Line $y=x+a \sqrt{2}$ is a tangent to the circle $x^2+y^2= a ^2$ at
  • A
    $\left(\frac{ a }{\sqrt{2}}, \frac{ a }{\sqrt{2}}\right)$
  • B
    $\left(-\frac{ a }{\sqrt{2}},-\frac{ a }{\sqrt{2}}\right)$
  • C
    $\left(\frac{ a }{\sqrt{2}},-\frac{ a }{\sqrt{2}}\right)$
  • ✓
    $\left(-\frac{ a }{\sqrt{2}}, \frac{ a }{\sqrt{2}}\right)$
Answer
Correct option: D.
$\left(-\frac{ a }{\sqrt{2}}, \frac{ a }{\sqrt{2}}\right)$
(D)
Suppose that the point be $( h , k )$.
Tangent at ( $h , k$ ) is
$\begin{array}{l}h x+k y=a^2 \text { and } x-y=-\sqrt{2} a \\\text { or } \frac{h}{l}=\frac{k}{-l}=\frac{a^2}{-\sqrt{2 a}} \text { or } h=-\frac{a}{\sqrt{2}}, k=\frac{a}{\sqrt{2}}\end{array}$
Therefore, point of contact is $\left(-\frac{ a }{\sqrt{2}}, \frac{ a }{\sqrt{2}}\right)$.
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MCQ 522 Marks
If the straight line y = mx + c touches the circle $x^2+y^2-4 y=0$, then the value of c will be
  • A
    $1+\sqrt{1+ m ^2}$
  • B
    $1-\sqrt{m^2+1}$
  • ✓
    $2\left(1+\sqrt{1+m^2}\right)$
  • D
    $2+\sqrt{1+m^2}$
Answer
Correct option: C.
$2\left(1+\sqrt{1+m^2}\right)$
(C)
Apply for tangency of line, $m x-y+ c =0$, centre being $(0,2)$ and radius $=2$$\left|\frac{-2+c}{\sqrt{1+m^2}}\right|=2$
$\begin{array}{l}\Rightarrow c ^2-4 c +4=4+4 m^2 \\ \Rightarrow c ^2-4 c -4 m^2=0 \\ \Rightarrow c =\frac{4 \pm \sqrt{16+16 m^2}}{2} \text { or } c =2\left(1+\sqrt{1+ m ^2}\right)\end{array}$
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MCQ 532 Marks
If the line ycos $\alpha=x \sin \alpha+a \cos \alpha$ be a tangent to the circle $x^2+y^2= a ^2$, then
  • A
    $\sin ^2 \alpha=1$
  • ✓
    $\cos ^2 \alpha=1$
  • C
    $\sin ^2 \alpha=a^2$
  • D
    $\cos ^2 \alpha=a^2$
Answer
Correct option: B.
$\cos ^2 \alpha=1$
(B)
The tangent is $y \cos \alpha=x \sin \alpha+\operatorname{acos} \alpha$
$\therefore y=x \tan \alpha+a$
Comparing with $y=m x+c$, we get
$m = x \tan \alpha, c = a$
It is a tangent to the circle $x^2+y^2= a ^2$,
If $c ^2= a ^2\left(1+ m ^2\right)$
i.e. $a^2=a^2\left(1+\tan ^2 \alpha\right)$
$\Rightarrow \sec ^2 \alpha=1$
$\Rightarrow \cos ^2 \alpha=1$
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MCQ 542 Marks
Which following lines is a tangent to the circle $x^2+y^2=25$ for all values of m?
  • A
    $y=m x+25 \sqrt{1+m^2}$
  • ✓
    $y=m x+5 \sqrt{1+m^2}$
  • C
    $y=m x+25 \sqrt{1-m^2}$
  • D
    $y=m x+5 \sqrt{1-m^2}$
Answer
Correct option: B.
$y=m x+5 \sqrt{1+m^2}$
(B)
Line $y= m x+ c$ is a tangent if
$c= \pm a \sqrt{1+m^2}$
$\therefore y=m x+5 \sqrt{1+m^2}$
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MCQ 552 Marks
The equations of the tangents to the circle $x^2+y^2=36$, which are inclined at an angle of $45^{\circ}$ to the X-axis are
  • A
    $x+y= \pm \sqrt{6}$
  • B
    $x=y \pm 3 \sqrt{2}$
  • ✓
    $y=x \pm 6 \sqrt{2}$
  • D
    $y=x \pm 2 \sqrt{6}$
Answer
Correct option: C.
$y=x \pm 6 \sqrt{2}$
(C)
$y= mx + c$ is a tangent, if
$c = \pm a \sqrt{1+ m ^2}$, where $m =\tan 45^{\circ}=1$
$\therefore$ The equation is $y=x \pm 6 \sqrt{2}$
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MCQ 562 Marks
The line 3x - 2y = k meets the circle $x^2+y^2=4 r^2$ at only point, if $k ^2$ is
  • A
    $20 r^2$
  • ✓
    $52 r^2$
  • C
    $\frac{52}{9} r^2$
  • D
    $\frac{20}{9} r^2$
Answer
Correct option: B.
$52 r^2$
(B)
$2 y=3 x- k$
$\therefore y=\frac{3}{2} x-\frac{ k }{2}$
Now $c ^2= a ^2\left(1+ m ^2\right)$
$\therefore \quad \frac{ k ^2}{4}=4 r ^2\left(1+\frac{9}{4}\right)$
$\therefore \quad k ^2=52 r ^2$
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MCQ 572 Marks
The line $\frac{x}{a}+\frac{y}{b}=1$ will touch the circle $x^2+y^2= c ^2$ if
  • A
    $\frac{1}{c}=\frac{1}{a}+\frac{1}{b}$
  • ✓
    $\frac{1}{ c ^2}=\frac{1}{ a ^2}+\frac{1}{b^2}$
  • C
    $c^2=a^2+b^2$
  • D
    c = a + b
Answer
Correct option: B.
$\frac{1}{ c ^2}=\frac{1}{ a ^2}+\frac{1}{b^2}$
(B)
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MCQ 582 Marks
If 5x - 12y + 10 = 0 and 12y - 5x + 16 = 0 are tangents of a circle, then radius of that circle is
  • A
    $4$
  • B
    $2$
  • ✓
    $1$
  • D
    $3$
Answer
Correct option: C.
$1$
(C)
The equation of the tangents are
$5 x-12 y+10=0$ and $5 x-12 y-16=0$
Hence, they are parallel to each other. The perpendicular distance between these two lines is the diameter of the circle
$2 r=\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|$
$c_1=10 ; c_2=-16 ; a=5 ; b=-12$
$\therefore \quad 2 r=\left|\frac{10-(-16)}{\sqrt{5^2+12^2}}\right|=\left|\frac{26}{13}\right|=2$
$\Rightarrow r=1$
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MCQ 592 Marks
The equation of the tangent to the circle $x^2+y^2+4 x-4 y+4=0$ which make equal intercepts on the positive co-ordinate axes is given by
  • A
    $x+y+2 \sqrt{2}=0$
  • ✓
    $x+y=2 \sqrt{2}$
  • C
    $x+y=2$
  • D
    $x+y=\sqrt{2}$
Answer
Correct option: B.
$x+y=2 \sqrt{2}$
(B)
Centre and radius of the circle $x^2+y^2+4 x-4 y+4=0$ are $(-2,2)$ and 2 respectively.
Let the equation of tangent be
$x+y+ c =0$.Then,
$\begin{array}{l}\left|\frac{-2+2+c}{\sqrt{2}}\right|=2 \\ \Rightarrow c= \pm 2 \sqrt{2}\end{array}$
But for positive intercepts, $c =-2 \sqrt{2}$
$\therefore$ The tangent is $x+y-2 \sqrt{2}=0$
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MCQ 602 Marks
If y + c = 0 is a tangent to the circle $x^2+y^2-6 x-2 y+1=0$ at (a, 4) then
  • A
    ac = 360
  • ✓
    ac = - 12
  • C
    a + c = 0
  • D
    4a = c
Answer
Correct option: B.
ac = - 12
(B)
Here, $g =-3, f =-1$
Equation of tangent at $(a, 4)$ is
$\begin{array}{l}a x+4 y-3(x+a)-(y+4)+1=0 \\\Rightarrow(a-3) x+3 y-3 a-4+1=0 \\\Rightarrow(a-3) x+3 y-3(a+1)=0 \\\Rightarrow \frac{(a-3)}{3} x+y-(a+1)=0\end{array}$
Comparing with $y+ c =0$, we get
$\frac{ a -3}{3}=0 \Rightarrow a =3$
$-(a+1)=c \Rightarrow c=-4$
$\therefore a c=-12$
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MCQ 612 Marks
If x + y = 2 is a tangent to $x^2+y^2=2$ then the equation of the tanget at the same point of contact to the circle $x^2+y^2+3 x+3 y-8=0$ is
  • A
    x - y = 6
  • B
    x + y = 6
  • ✓
    x + y = 2
  • D
    x - y = 2
Answer
Correct option: C.
x + y = 2
(C)
Equation of the tangent to $x^2+y^2=2$ at $( h , k )$ is $h x+ k y=2$
By comparing $h x+ k y=2$ with $x+y=2$,
point of contact is $(1,1)$.
Equation of the tangent to $x^2+y^2+3 x+3 y-8=0$ at $(1,1)$ is
$\begin{array}{l}x(1)+y(1)+\frac{3}{2}(x+1)+\frac{3}{2}(y+1)-8=0 \\\Rightarrow 5 x+5 y-10=0 \\\Rightarrow x+y=2\end{array}$
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MCQ 622 Marks
The X-axis touches the circle whose centre is (0, 1) The equation of the tangent to the circle at (1, 1) is
  • A
    x + 2 = 0
  • ✓
    x - 1 = 0
  • C
    x - 2 = 0
  • D
    x + y - 2 = 0
Answer
Correct option: B.
x - 1 = 0
(B)
The equation of the circle with centre $(0,1)$ is
$x^2+(y-1)^2=a^2$
It passes through the point $(1,1)$.
$\therefore$ $1^2+(1-1)^2=a^2$
$\Rightarrow$ radius is 1 .
The equation of the circle is $x^2+y^2-2 y=0$.
$\therefore$ The equation of the tangent at $(1,1)$ is
$\begin{array}{l}x+y-y-1=0 \\\Rightarrow x-1=0\end{array}$
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MCQ 632 Marks
If the centre of a circle is (-6, 8) and it passes through the origin, then equation to its tangent at origin, is
  • A
    2y = x
  • ✓
    4y = 3x
  • C
    3y = 4x
  • D
    3x + 4y = 0
Answer
Correct option: B.
4y = 3x
(B)
Centre $(-6,8)$, radius $=\sqrt{6^2+8^2}=10$
$\therefore$ Equation of circle is $x^2+y^2+12 x-16 y=0$
$\therefore$ Equation of tangent at $(0,0)$ is
$6 x-8 y=0 \Rightarrow 3 x=4 y$
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MCQ 642 Marks
The area of triangle formed by the tangent normal drawn at $(1, \sqrt{3})$ to the circle $x^2+y^2=4$ and positive X-axis, is
  • ✓
    $2 \sqrt{3}$
  • B
    $\sqrt{3}$
  • C
    $4 \sqrt{3}$
  • D
    $3 \sqrt{3}$
Answer
Correct option: A.
$2 \sqrt{3}$
(A)
Equation of the tangent at $(1, \sqrt{3})$ is
$x+\sqrt{3} y-4=0$
Image
$PM=\sqrt{3} \text { and } OR=4$
Hence, the required area $=\frac{1}{2} \times 4 \times \sqrt{3}=2 \sqrt{3}$
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MCQ 652 Marks
Equation of tangent to the circle $x^2+y^2=10$ at the point with abscissa 1 is
  • ✓
    $x \pm 3 y=10$
  • B
    $3 x \pm y=10$
  • C
    $3 x \pm 3 y=10$
  • D
    $x-y=3$
Answer
Correct option: A.
$x \pm 3 y=10$
(A)
Abscissa $=1$
Hence, given equation of circle reduces to
$y^2=9$
$\Rightarrow y= \pm 3$
$\therefore$ Equation of tangent at
$(1, \pm 3)$ to $x^2+y^2=10$ is $x(1)+y( \pm 3)=10$
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MCQ 662 Marks
The equation of the tangent to the circle $x^2+y^2=50$ at the point, where the line x - 7 = 0 meets the circle, is
  • A
    7x + y = 50
  • B
    x + 7y = 50
  • C
    $x \pm 7 y=50$
  • ✓
    $7 x \pm y=50$
Answer
Correct option: D.
$7 x \pm y=50$
(D)
The equation of the tangent to the circle $x^2+y^2=50$ is $x x_1+y y_1=50$
The circle meets the line $x-7$.
$\therefore (7)^2+y_1^2=50$
$\Rightarrow y_1^2=50-49=1$
$\Rightarrow y_1= \pm 1$
$\therefore$ Equation of the tangent is $7 x \pm y=50$
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MCQ 672 Marks
Let the line segment joining the centres of the circles $x^2-2 x+y^2=0$ and $x^2+y^2+4 x+8 y+16=0$ intersect the another circle at P and Q respectively. Then the equation of the circle with PQ as its diameter is
  • A
    $5 x^2+5 y^2-2 x-16 y+8=0$
  • B
    $5 x^2+5 y^2-8 x-24 y+27=0$
  • C
    $5 x^2+5 y^2+8 x+24 y+27=0$
  • ✓
    $5 x^2+5 y^2+2 x+16 y+8=0$
Answer
Correct option: D.
$5 x^2+5 y^2+2 x+16 y+8=0$
(D)
The centres of two circles are $C_1(1,0)$ and $C_2(-2,-4)$ and their radii are 1 and 2 units respectively.
Let C be the centre of the required circle.
Then, $C P=C Q=1$.
$\therefore$ $C C_1=2$ and $C C_2=3$.
Clearly, C divides $C _1 C _2$ in the ratio $2: 3$.
Therefore, coordinates of C are
$\left(\frac{-4+3}{2+3}, \frac{-8+0}{2+3}\right)=\left(-\frac{1}{5},-\frac{8}{5}\right) .$
Image

Hence, equation of the required circle is
$\begin{array}{l}\left(x+\frac{1}{5}\right)^2+\left(y+\frac{8}{5}\right)^2=1^2 \\\Rightarrow 5 x^{2}+5 y^{2}+2 x+16 y+8=0\end{array}$
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MCQ 682 Marks
The equation of the circle which passes through the points of intersection of the circles $x^2+y^2-6 x=0$ and $x^2+y^2-6 y=0$ and has its centre at $\left(\frac{3}{2}, \frac{3}{2}\right)$ is
  • A
    $x^2+y^2+3 x+3 y+9=0$
  • B
    $x^2+y^2+3 x+3 y=0$
  • ✓
    $x^2+y^2-3 x-3 y=0$
  • D
    $x^2+y^2-3 x-3 y+9=0$
Answer
Correct option: C.
$x^2+y^2-3 x-3 y=0$
(C)
$C_1: x^2+y^2-6 x=0$ ....(i)
$C _2: x^2+y^2-6 y=0$ ....(ii)
Solving (i) and (ii), we get
$x=y$ ....(iii)
Substituting (iii) in (i), we get
$y=3$
$\therefore x=3$
Point on circle is $P (3,3)$ and
$\begin{aligned} & \text { centre }=\left(\frac{3}{2}, \frac{3}{2}\right) \\ \therefore & \text { Radius }=\sqrt{\left(3-\frac{3}{2}\right)^2+\left(3-\frac{3}{2}\right)^2}=\frac{3}{\sqrt{2}}\end{aligned}$
$\therefore$ equation of the circle is
$\begin{array}{l}\left(x-\frac{3}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\frac{9}{2} \\\Rightarrow x^2+y^2-3 x-3 y=0\end{array}$
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MCQ 692 Marks
If one diameters of the circle $x^2+y^2-2 x-6 y+6=0$ is a chord to the circle with centre (2, 1), then the radius of the bigger circle is
  • A
    $6$
  • B
    $4$
  • C
    $2$
  • ✓
    $3$
Answer
Correct option: D.
$3$
(D)
The centre of the given circle is $C_1(1,3)$ and
radius $=\sqrt{(-1)^2+(-3)^2-6}$
$=\sqrt{1+9-6}$
$=2$
$C _1 C _2=\sqrt{(2-1)^2+(1-3)^2}$
$=\sqrt{1+4}=\sqrt{5}$
$\therefore$ Radius of bigger circle $=\sqrt{(\sqrt{5})^2+2^2}$
$=\sqrt{5+4}$
$=3$
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MCQ 702 Marks
If one of given diameters of the circle, given by the equation, $x^2+y^2-4 x+6 y-12=0$, is a chord of a circle S, whose centre is at (-3, 2), then the radius of S is
  • ✓
    $5 \sqrt{3}$
  • B
    $5$
  • C
    $10$
  • D
    $5 \sqrt{2}$
Answer
Correct option: A.
$5 \sqrt{3}$
(A)
Image
The centre of the given circle is $C _1(2,-3)$ and radius
$\begin{array}{l}=\sqrt{(-2)^2+3^2-(-12)} \\=5 \\C_1 C_2=\sqrt{(-3-2)^2+(2+3)^2} \\\quad=\sqrt{25+25} \\\quad=\sqrt{50}\end{array}$
$\therefore$ Radius of S is $C _2 A=\sqrt{(\sqrt{50})^2+5^2}$
$\begin{array}{l}=\sqrt{75} \\=5 \sqrt{3}\end{array}$
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MCQ 712 Marks
If one of the diameters of the curve $x^2+y^2-4 x-6 y+9=0$ is a chord of a circle with centre (1, 1), then the radius of this circle is
  • ✓
    $3$
  • B
    $2$
  • C
    $\sqrt{2}$
  • D
    $1$
Answer
Correct option: A.
$3$
(A)
Given equation of circle is
$x^2+y^2-4 x-6 y+9=0$
$\Rightarrow x^2-4 x+4+y^2-6 y+9-4=0$
$\Rightarrow(x-2)^2+(y-3)^2=4$
$\therefore$ centre $=(2,3)$, radius $=2$
The diameter of this circle is a chord of circle with centre $O (1,1)$.
Image
$OP =\sqrt{(3-1)^2+(2-1)^2}=\sqrt{5}$
$QP =2$
$\therefore r^2=(\sqrt{5})^2+2^2 \Rightarrow r=3$
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MCQ 722 Marks
For all values of $\theta$, the locus of the point of intersection of the lines $x \cos \theta+y \sin \theta= a$ and $x \sin \theta-y \cos \theta=b$ is
  • A
    an ellipse
  • ✓
    a circle
  • C
    a parabola
  • D
    a hyperbola
Answer
Correct option: B.
a circle
(B)
The point of intersection is
$x= a \cos \theta+ b \sin \theta$
$y= a \sin \theta- b \cos \theta$
$\therefore \quad x^2+y^2= a ^2+ b ^2$
Hence, it is equation of a circle.
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MCQ 732 Marks
Radius parametric equation represented by $x=2 a\left(\frac{1-t^2}{1+t^2}\right), y=\frac{4 a t}{1+t^2}$ is
  • A
    $a$
  • B
    $a ^2$
  • C
    3a
  • ✓
    2a
Answer
Correct option: D.
2a
(D)
$x=2 a\left(\frac{1-t^2}{1+t^2}\right)$ ...(i)
$y=\frac{4 at }{1+ t ^2}$ ...(ii)
Squaring and adding (i) and (ii), we get
$x^2+y^2=4 a ^2 \cdot \frac{\left(1- t ^2\right)^2}{\left(1+ t ^2\right)^2}+\frac{16 a ^2 t ^2}{\left(1+ t ^2\right)^2}$
$=\frac{4 a^2}{\left(1+t^2\right)^2}\left[1-2 t^2+t^4+4 t^2\right]$
$=\frac{4 a^2}{\left(1+t^2\right)^2}\left(1+t^2\right)^2$
$\therefore \quad x^2+y^2=(2 a)^2$
$\therefore \quad$ Radius $=2 a$
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MCQ 742 Marks
For what value of k, the points (0, 0), (1, 3), (2, 4) and (k, 3) are con-cyclic?
  • A
    $2$
  • ✓
    $1$
  • C
    $4$
  • D
    $5$
Answer
Correct option: B.
$1$
(B)
The equation of circle through points $(0,0)$, $(1,3)$ and $(2,4)$ is
$x^2+y^2-10 x=0$
Point ( $k, 3$ ) will be on the circle, if
$\begin{array}{l}k^2+9-10 k=0 \\\Rightarrow k^2-10 k+9=0 \\\Rightarrow k^2-9 k-k+9=0 \\\Rightarrow(k-1)(k-9)=0 \\\Rightarrow k=1 \text { or } k=9\end{array}$
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MCQ 752 Marks
If a circle passes through the points (0, 0), (a, 0), (0, b), then its centre is
  • A
    (a, b)
  • B
    (b, a)
  • ✓
    $\left(\frac{a}{2}, \frac{b}{2}\right)$
  • D
    $\left(\frac{b}{2},-\frac{a}{2}\right)$
Answer
Correct option: C.
$\left(\frac{a}{2}, \frac{b}{2}\right)$
(C)
Let the equation of circle be
$x^2+y^2+2 g x+2 f y+c=0$
Now on passing through the given points, we get three equations
$c=0$ ...(i)
$a^2+2 g a+c=0$....(ii)
$b^2+2 f b+c=0$ ....(iii)
Solving equations (i), (ii) and (iii), we get
$g=-\frac{a}{2}, f=-\frac{b}{2}$
Hence, the centre is $\left(\frac{ a }{2}, \frac{b}{2}\right)$.
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MCQ 762 Marks
The locus of the centre of the circle which cuts off intercepts of length 2a and 2b from X-axis and Y-axis respectively, is
  • A
    $x+y= a + b$
  • B
    $x^2+y^2=a^2+b^2$
  • ✓
    $x^2-y^2=a^2-b^2$
  • D
    $x^2+y^2=a^2-b^2$
Answer
Correct option: C.
$x^2-y^2=a^2-b^2$
(C)
Since X -intercept $=2 a$
$\therefore 2 \sqrt{g^2-c}=2 a$$\quad\ldots(i)$
Also, Y -intercept $=2 b$
$\therefore 2 \sqrt{f^2-c}=2 b$$\quad\ldots(ii)$
On squaring (i) and (ii) and then subtracting (ii) from (i), we get
$g^2-f^2=a^2-b^2$
Hence, the locus is
$x^2-y^2=a^2-b^2$
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MCQ 772 Marks
The equation of the circle passing through the point (2, 1) and touching Y-axis at the origin is
  • A
    $x^2+y^2-5 x=0$
  • ✓
    $2 x^2+2 y^2-5 x=0$
  • C
    $x^2+y^2+5 x=0$
  • D
    None of these
Answer
Correct option: B.
$2 x^2+2 y^2-5 x=0$
(B)
We have the equation of circle
$x^2+y^2+2 g x+2 f y+c=0$
But it passes through $(0,0)$ and $(2,1)$.
$\therefore \quad c=0$
$5+4 g+2 f=0$ ....(i)
Also $\sqrt{ g ^2+ f ^2- c }=| g |$
$\Rightarrow f =0$ $\ldots [\because c=0]$
$\therefore g=-\frac{5}{4}$ ....[From (i)]
Hence, the equation will be $2 x^2+2 y^2-5 x=0$.
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MCQ 782 Marks
A circle is concentric with the circle $x^2+y^2-6 x+12 y+15=0$ and has area double of its area. The equation of the circle is
  • ✓
    $x^2+y^2-6 x+12 y-15=0$
  • B
    $x^2+y^2-6 x+12 y+15=0$
  • C
    $x^2+y^2-6 x+12 y+45=0$
  • D
    $x^2+y^2-6 x+12 y-45=0$
Answer
Correct option: A.
$x^2+y^2-6 x+12 y-15=0$
(A)
Equation of circle concentric to given circle is
$x^2+y^2-6 x+12 y+k=0$
Since area of required circle $=2$ (area of given circle)$\begin{array}{l}\Rightarrow \sqrt{9+36-k}=\sqrt{2} \sqrt{9+36-15} \\\Rightarrow 45-k=60 \\\Rightarrow k=-15\end{array}$
Hence, the required equation of circle is
$x^2+y^2-6 x+12 y-15=0$
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MCQ 792 Marks
The equation of the circle whose radius is 5 and which touches the circle $x^2+y^2-2 x-4 y-20=0$ externally at the point (5, 5), is
  • A
    $x^2+y^2-18 x-16 y-120=0$
  • ✓
    $x^2+y^2-18 x-16 y+120=0$
  • C
    $x^2+y^2+18 x+16 y-120=0$
  • D
    $x^2+y^2+18 x-16 y+120=0$
Answer
Correct option: B.
$x^2+y^2-18 x-16 y+120=0$
(B)
Let the centre of the required circle be $\left(x_1, y_1\right)$. Centre of given circle is $(1,2)$ and
$r =\sqrt{1+4+20}=5$
$\therefore$ radii of both circles are same.
$\therefore$ Point of contact $(5,5)$ is the mid point of the line joining the centres of both circles.
$\begin{array}{l} \therefore \frac{x_1+1}{2}=5 \text { and } \frac{y_1+2}{2}=5 \\\Rightarrow x_1=9, y_1=8\end{array}$
Hence, the required equation is
$(x-9)^2+(y-8)^2=25$
$\Rightarrow x^2+y^2-18 x-16 y+120=0$
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MCQ 802 Marks
If the circles $x^2+y^2+2 \lambda x+2=0$ and $x^2+y^2+4 y+2=0$ touch each other, then $\lambda=$
Answer
Correct option: B.
$\pm 2$
(B)
The centres of two circles are $C_1(-\lambda, 0)$ and $C _2(0,-2)$ and their radii are $\sqrt{\lambda^2-2}$ and $\sqrt{2}$.
The given circles touches each other, if
$\sqrt{\lambda^2+4}=\sqrt{\lambda^2-2}+\sqrt{2}$
$\Rightarrow \lambda^2+4=\lambda^2-2+2+2 \sqrt{2} \sqrt{\lambda^2-2}$
$\Rightarrow \sqrt{2}=\sqrt{\lambda^2-2}$
$\Rightarrow \lambda^2=4$
$\Rightarrow \lambda= \pm 2$
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MCQ 812 Marks
The equation of a circle of radius 5 which lies within the circle $x^2+y^2+14 x+10 y-26=0$ and touches it at the point (-1, 3) is
  • ✓
    $x^2+y^2+8 x+2 y-8=0$
  • B
    $x^2+y^2+10 x+2 y+1=0$
  • C
    $x^2+y^2+8 x+4 y-4=0$
  • D
    $x^2+y^2+8 x+6 y=0$
Answer
Correct option: A.
$x^2+y^2+8 x+2 y-8=0$
(A) Consider option (A),
$x^2+y^2+8 x+2 y-8=0$
Point $(-1,3)$ is common to both circle and lies on above circle also.
Since point $(-1,3)$ satisfies the equation of circle in option (A).
$\therefore$ correct answer is option (A).
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MCQ 822 Marks
A circle $x^2+y^2+2 g x+2 f y+ c =0$ passing through (4, -2) is concentric to the circle $x^2+y^2-2 x+4 y+20=0$ then the value of c will be
  • ✓
    -4
  • B
    4
  • C
    $0$
  • D
    $1$
Answer
Correct option: A.
-4
(A)
Circle $x^2+y^2+2 gr +2 fy + c =0$
is concentric with $x^2+y^2-2 x+4 y+20=0$.
$\therefore$ centre is $(1,-2)$ and
radius $=\sqrt{(4-1)^2+(-2+2)^2}=\sqrt{3^2+0^2}=3$
Also, $r=\sqrt{g^2+f^2-c}$
$\therefore 3=\sqrt{(-1)^2+(2)^2-c}$
$\therefore 9=1+4-c$
$\therefore c=-4$
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MCQ 832 Marks
The equation of the circle concentric with the circle $x^2+y^2+8 x+10 y-7=0$ and passing through the centre of the circle $x^2+y^2-4 x-6 y=0$ is
  • A
    $x^2+y^2+8 x+10 y+59=0$
  • ✓
    $x^2+y^2+8 x+10 y-59=0$
  • C
    $x^2+y^2-4 x-6 y+87=0$
  • D
    $x^2+y^2-4 x-6 y-87=0$
Answer
Correct option: B.
$x^2+y^2+8 x+10 y-59=0$
(B)
Centre of the required circle is $(-4,-5)$ and it passes through $(2,3)$.
$\therefore$ $\text { Radius }=\sqrt{(-4-2)^2+(-5-3)^2}=10$
$\therefore$ Equation of the required circle is
$(x+4)^2+(y+5)^2=(10)^2$
$\Rightarrow x^2+y^2+8 x+10 y-59=0$
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MCQ 842 Marks
The equation of the circle with centre at (1, -2) and passing through the centre of the given circle $x^2+y^2+2 y-3=0$, is
  • ✓
    $x^2+y^2-2 x+4 y+3=0$
  • B
    $x^2+y^2-2 x+4 y-3=0$
  • C
    $x^2+y^2+2 x-4 y-3=0$
  • D
    $x^2+y^2+2 x-4 y+3=0$
Answer
Correct option: A.
$x^2+y^2-2 x+4 y+3=0$
(A)
Centre of the given circle is $(0,-1)$.
$\therefore$ the required circle passes through $(0,-1)$.
$\therefore$ $r=\sqrt{(0-1)^2+(-1+2)^2}=\sqrt{ } 2$
Hence, the required equation is
$(x-1)^2+(y+2)^2=(\sqrt{2})^2$
$\Rightarrow x^2+y^2-2 x+4 y+3=0$
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MCQ 852 Marks
The equation of the circle concentric with the circle $x^2+y^2-4 x-6 y-3=0$ and touching Y-axis is
  • A
    $x^2+y^2-4 x-6 y-9=0$
  • ✓
    $x^2+y^2-4 x-6 y+9=0$
  • C
    $x^2+y^2-4 x-6 y+3=0$
  • D
    $x^2+y^2-4 x-6 y-3=0$
Answer
Correct option: B.
$x^2+y^2-4 x-6 y+9=0$
(B)
Centre of the circle
$x^2+y^2-4 x-6 y-3=0 \text { is } C(2,3)$
Since, it touches the Y -axis
$\therefore$ $r-2$
Hence required equation of the circle is
$(x-2)^2+(y-3)^2=2^2$
$\Rightarrow x^2+y^2-4 x-6 y+9=0$
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MCQ 862 Marks
A variable circle passes through the fixed point A(p, q) and touches X-axis. The locus of the other end of diameter through A is
  • ✓
    $(x-p)^2=4 q y$
  • B
    $(x-q)^2=4 p y$
  • C
    $(y-p)^2=4 q x$
  • D
    $(y-q)^2=4 p x$
Answer
Correct option: A.
$(x-p)^2=4 q y$
(A)
Let another end of the diameter be ( $h , k$ ).
Since centre is the midpoint of the diameter.
$\therefore \quad$ Centre $=\left(\frac{ p + h }{2}, \frac{ q + k }{2}\right)$
Since the circle touches X -axis,
radius $=\left|\frac{q+k}{2}\right|$
$\Rightarrow \sqrt{( h - p )^2+( k - q )^2}=2\left|\frac{ q + k }{2}\right|$
$\Rightarrow( h - p )^2+( k - q )^2=( q + k )^2$
$\Rightarrow( h - p )^2=( k + q )^2-( k - q )^2$
$\Rightarrow( h - p )^2=4 k y$
$\therefore $ Locus of $( h , k )$ is $(x- p )^2=4 q y$
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MCQ 872 Marks
The point diametrically opposite to the point P(1, 0) on the circle $x^2+y^2+2 x+4 y-3=0$ is
  • A
    (-3, 4)
  • ✓
    (-3, -4)
  • C
    (3, 4)
  • D
    (3, -4)
Answer
Correct option: B.
(-3, -4)
(B)
Image
Let $A (x, y)$ be the required point.
Given equation of circle is
$x^2+y^2+2 x+4 y-3=0$
$\therefore$ Centre $=(-1,-2)$
Since C is the midpoint of AP.
$\therefore$ $A=(-3,-4)$
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MCQ 882 Marks
If one end of a diameter of the circle $x^2+y^2-4 x-6 y+11=0$ be (3, 4), then the other end is
  • A
    (0, 0)
  • B
    (1, 1)
  • ✓
    (1, 2)
  • D
    (2, 1)
Answer
Correct option: C.
(1, 2)
(C)
Let another end of the diameter be $(x, y)$.
Centre of the given circle is $(2,3)$.
Since centre is the midpoint of the diameter.
$\therefore 2=\frac{3+x}{2}, 3=\frac{4+y}{2}$
$\Rightarrow x=1, y=2$
$\Rightarrow(x, y)=(1,2)$
Alternate Method:
Here, $\left(x_1, y_1\right)=(3,4)$
$\therefore$ $(x, y)-(-(3-4),-(4-6))$...[Using Shortcut 6]$=(1,2)$
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MCQ 892 Marks
If the equation $\frac{K(x+1)^2}{3}+\frac{(y+2)^2}{4}=1$ represents a circle, then K =
  • ✓
    $\frac{3}{4}$
  • B
    $1$
  • C
    $\frac{4}{3}$
  • D
    12
Answer
Correct option: A.
$\frac{3}{4}$
(A)
The given equation represents a circle, if coeff. of $x^2=\operatorname{coeff}$, of $y^2$
After solving the given equation , we get
$\frac{K}{3}=\frac{1}{4} \Rightarrow K=\frac{3}{4}$
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MCQ 902 Marks
The equation of the circle passing through the point (1, 0) and (0, 1) and having the smallest radius is
  • A
    $x^2+y^2-2 x-2 y+1=0$
  • ✓
    $x^2+y^2-x-y=0$
  • C
    $x^2+y^2+2 x+2 y-7=0$
  • D
    $x^2+y^2+x+y-2=0$
Answer
Correct option: B.
$x^2+y^2-x-y=0$
(B)
Circle whose diametric end points are $(1,0)$ and $(0,1)$ will be of smallest radius.
$\therefore$ By using diameter form, equation of circle is
$(x-1)(x-0)+(y-0)(y-1)=0$
$\Rightarrow x^2+y^2-x-y=0$
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MCQ 912 Marks
The equation of the circle whose centre is $(1,-3)$ and which touches the line $2 x-y-4=0$ is
  • ✓
    $5 x^2+5 y^2-10 x+30 y+49=0$
  • B
    $5 x^2+5 y^2+10 x-30 y+49=0$
  • C
    $5 x^2+5 y^2-10 x+30 y-49=0$
  • D
    $5 x^2+5 y^2-10 x-30 y+49=0$
Answer
Correct option: A.
$5 x^2+5 y^2-10 x+30 y+49=0$
(A)
$\text { Radius of circle }=\left|\frac{2(1)-1(-3)-4}{\sqrt{4+1}}\right|=\frac{1}{\sqrt{5}}$
$\therefore$ Equation is $(x-1)^2+(y+3)^2=\left(\frac{1}{\sqrt{5}}\right)^2$
$\Rightarrow x^2+y^2-2 x+6 y+10=\frac{1}{5}$
$\Rightarrow 5 x^2+5 y^2-10 x+30 y+49=0$
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MCQ 922 Marks
The equation of a circle which touches both axes and the line 3x -4y + 8 = 0 and whos centre lies in the third quadrant is
  • A
    $x^2+y^2-4 x+4 y-4=0$
  • B
    $x^2+y^2-4 x+4 y+4=0$
  • ✓
    $x^2+y^2+4 x+4 y+4=0$
  • D
    $x^2+y^2-4 x-4 y-4=0$
Answer
Correct option: C.
$x^2+y^2+4 x+4 y+4=0$
(C)
The equation of circle in third quadrant touching the coordinate axes with centre $(- a ,- a )$ and radius ' a ' is
$x^2+y^2+2 a x+2 a y+a^2=0$  ...(i)
Since, line $3 x-4 y+8=0$ touches the circle
$\therefore$ perpendiular distance from centre of the circle to the line $=$ radius
$\therefore$ $\left|\frac{3(-a)-4(-a)+8}{\sqrt{9+16}}\right|=a$
$\Rightarrow a=2$
Substituting $a =2$ in equation (i), we get
$x^2+y^2+4 x+4 y+4=0$
This is the required equation of the circle
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MCQ 932 Marks
The equation of the circle whose centre is (3, -1) and which cuts off a chord of length 6 on the line 2x - 5y + 18 = 0 is
  • A
    $(x-3)^2+(y+1)^2=38$
  • B
    $(x+3)^2+(y-1)^2=38$
  • C
    $(x-3)^2+(y+1)^2=\sqrt{38}$
  • D
    None of these
Answer
(A)
Image
Let AB be the chord cut off by the circle on the line $2 x-5 y+18=0$.
Let CD be the perpendicular drawn from centre $(3,-1)$ to AB .
$\begin{array}{ll}\therefore & CD=\left|\frac{2(3)-5(-1)+18}{\sqrt{2^2+(-5)^2}}\right|=\sqrt{29} \\& \text { and } AD=3 \\ \therefore & CA^2=AD^2\left|CD^2=3^2\right|(\sqrt{29})^2=38\end{array}$
∴ the equation of the circle is
$(x-3)^2+(y+1)^2=38$
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MCQ 942 Marks
Equation of the circle which touches the lines x = 0, y = 0 and 3x + 4y = 4 is
  • A
    $x^2-4 x+y^2+4 y+4=0$
  • ✓
    $x^2-4 x+y^2-4 y+4=0$
  • C
    $x^2+4 x+y^2+4 y+4=0$
  • D
    $x^2+4 x+y^2-4 y+4=0$
Answer
Correct option: B.
$x^2-4 x+y^2-4 y+4=0$
(B)
Let centre of circle be ( $h , k$ ).
Since it touches both axes, therefore $h = k = a$
Hence, equation can be $(x-a)^2+(y-a)^2=a^2$
But it also touches the line $3 x+4 y=4$
$\therefore\left|\frac{3 a+4 a-4}{\sqrt{9+16}}\right|=a$
$\Rightarrow a =2$
Hence, the required equation of circle is
$(x-2)^2+(y-2)^2=2^2$
$\Rightarrow x^2+y^2-4 x-4 y+4=0$
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MCQ 952 Marks
The equation of the circle which passes throug the points (2, 3) and (4, 5) and the centre lies on the straight line y - 4x + 3 = 0, is
  • A
    $x^2+y^2+4 x-10 y+25=0$
  • ✓
    $x^2+y^2-4 x-10 y+25=0$
  • C
    $x^2+y^2-4 x-10 y+16=0$
  • D
    $x^2+y^2-14 y+8=0$
Answer
Correct option: B.
$x^2+y^2-4 x-10 y+25=0$
(B)
Let centre be ( $h , k$ ). Then,
$\sqrt{( h -2)^2+( k -3)^2}=\sqrt{( h -4)^2+( k -5)^2}$
$\Rightarrow-4 h+4-6 k+9=-8 h+16-10 k+25$
$\Rightarrow 4 h+4 k-28=0$
$\Rightarrow h + k -7=0$   ...(i)
Since, centre lies on the given line.
$\therefore$ $k-4 h+3=0$....(ii)
Solving (i) and (ii), we get $( h , k )=(2,5)$
$\therefore$ centre is $(2,5)$ and
$\text { radius }=\sqrt{(2-2)^2+(5-3)^2}=2$
$\therefore$ the required equation of the circle is$\begin{array}{l}(x-2)^2+(y-5)^2=(2)^2 \\\Rightarrow x^2+y^2-4 x-10 y+25=0\end{array}$
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MCQ 962 Marks
A circle has radius 3 units and its centre lies of the line y = x - 1. Then the equation of the circle if it passes through point (7, 3), is
  • ✓
    $x^2+y^2-8 x-6 y+16=0$
  • B
    $x^2+y^2+8 x+6 y+16=0$
  • C
    $x^2+y^2-8 x-6 y-16=0$
  • D
    $x^2+y^2+8 x-6 y-16=0$
Answer
Correct option: A.
$x^2+y^2-8 x-6 y+16=0$
(A)
Let its centre be ( $h , k$ ), then
$h-k=1$....(i)
Also, radius $a =3$
$\therefore$ Equation of the circle is$(x-h)^2+(y-k)^2=9$
Also, it passes through $(7,3)$
$\text { i.e., }(7-h)^2+(3-k)^2=9$.....(ii)
From (i) and (ii), we get
$h=4, k=3$
$\therefore$ Equation is $x^2+y^2-8 x-6 y+16=0$
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MCQ 972 Marks
The lines 2x - 3y = 5 and 3x - 4y = 7 are the diameters of a circle of area 154 square units The equation of the circle is
  • A
    $x^2+y^2+2 x-2 y=62$
  • ✓
    $x^2+y^2-2 x+2 y=47$
  • C
    $x^2+y^2+2 x-2 y=47$
  • D
    $x^2+y^2-2 x+2 y=62$
Answer
Correct option: B.
$x^2+y^2-2 x+2 y=47$
(B)
Centre of circle $=$ Point of intersection of diameters $=(1,-1)$
Now, area $=154$
$\Rightarrow \pi r^2=154\Rightarrow r=7$
Hence, the equation of required circle is
$(x-1)^2+(y+1)^2=7^2$
$\Rightarrow x^2+y^2-2 x+2 y=47$
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MCQ 982 Marks
If the lines 2x + 3y + 1 = 0 and 3x -y - 4 = 0 lie along diameters of a circle of circumferenc $10 \pi$ , then the equation of the circle is
  • A
    $x^2+y^2+2 x-2 y-23=0$
  • B
    $x^2+y^2-2 x-2 y-23=0$
  • C
    $x^2+y^2+2 x-2 y+23=0$
  • ✓
    $x^2+y^2-2 x+2 y-23=0$
Answer
Correct option: D.
$x^2+y^2-2 x+2 y-23=0$
(D)
Since the centre always lies on the diameter.
Solving $2 x+3 y+1=0$ and $3 x-y-4=0$, the co-ordinates of the centre are $(1,-1)$.
Given, circumference $=10 \pi$
$\therefore \quad 2 \pi r=10 \pi \Rightarrow r=5$
∴ the equation of the circle is
$(x-1)^2+(y+1)^2=5^2$
$\Rightarrow x^2+y^2-2 x+2 y-23=0$
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MCQ 992 Marks
The equation of the circle whose diameter lies on 2x+3y=3 and 16x-y=4 which passes through (4, 6) is
  • ✓
    $5\left(x^2+y^2\right)-3 x-8 y=200$
  • B
    $x^2+y^2-4 x-8 y=200$
  • C
    $5\left(x^2+y^2\right)-4 x=200$
  • D
    $x^2+y^2=40$
Answer
Correct option: A.
$5\left(x^2+y^2\right)-3 x-8 y=200$
(A)
Since the centre always lies on the diameter.
Solving $2 x+3 y=3$ and $16 x-y=4$,
we get co-ordinates of the centre $=\left(\frac{3}{10}, \frac{4}{5}\right)$.
The circle passes through $(4,6)$.
$\therefore \quad r^2=\left(4-\frac{3}{10}\right)^2+\left(6-\frac{4}{5}\right)^2$
$=\left(\frac{37}{10}\right)^2+\left(\frac{26}{5}\right)^2=\frac{4073}{100}$
∴ the equation of the circle is
$\left(x-\frac{3}{10}\right)^2+\left(y-\frac{4}{5}\right)^2=\frac{4073}{100}$
$\Rightarrow 100 x^2+100 y^2-60 x-160 y=4000$
$\Rightarrow 5\left(x^2+y^2\right)-3 x-8 y=200$
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MCQ 1002 Marks
If the lines x + y = 6 and x + 2y = 4 be diameters of the circle whose diameter is 20, then the equation of the circle is
  • ✓
    $x^2+y^2-16 x+4 y-32=0$
  • B
    $x^2+y^2+16 x+4 y-32=0$
  • C
    $x^2+y^2+16 x+4 y+32=0$
  • D
    $x^2+y^2+16 x-4 y+32=0$
Answer
Correct option: A.
$x^2+y^2-16 x+4 y-32=0$
(A)
Here, $r =10$ (radius)
Centre will be the point of intersection of the diameters, i.e., $(8,-2)$.
Hence, required equation is
$\begin{array}{l}(x-8)^2+(y+2)^2=10^2 \\\Rightarrow x^2+y^2-16 x+4 y-32=0\end{array}$
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MCQ 1012 Marks
The equation of the circle having centre (1, -2) and passing through the point of intersection of lines 3x + y = 14, 2x + 5y = 18 is
  • ✓
    $x^2+y^2-2 x+4 y-20=0$
  • B
    $x^2+y^2-2 x-4 y-20=0$
  • C
    $x^2+y^2+2 x-4 y-20=0$
  • D
    $x^2+y^2+2 x+4 y-20=0$
Answer
Correct option: A.
$x^2+y^2-2 x+4 y-20=0$
(A)
The point of intersection of $3 x+y-14=0$ and $2 x+5 y-18=0$ is $(4,2)$.
Centre of the circle is $(1,2)$.
$\therefore$ radius $=\sqrt{(4-1)^2+(2+2)^2}=5$
∴ the equation of the circle is
$(x-1)^2+(y+2)^2=5^2$
$\therefore x^2+y^2-2 x+4 y-20=0$
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MCQ 1022 Marks
The length of the diameter of the circle which touches the X-axis at the point (1, 0) and passes through the point (2, 3) is
  • A
    $\frac{10}{3}$
  • B
    $\frac{3}{5}$
  • C
    $\frac{6}{5}$
  • D
    $\frac{5}{3}$
Answer
(A)
Image
Since the circle touches X -axis at $(1,0)$.
$\therefore $ centre of the circle is $(1, k)$ and radius $=k$
$\therefore $ equation of the circle is $(2-1)^2+(3-k)^2=k^2$
$\Rightarrow 1+ k ^2-6 k +9= k ^2$
$\Rightarrow k =\frac{5}{3}$
$\therefore$ diameter $=2 k =\frac{10}{3}$
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MCQ 1032 Marks
The circle passing through the point (-1, 0) and touching the Y-axis at (0, 2) also passes through the point
  • A
    $\left(-\frac{3}{2}, 0\right)$
  • B
    $\left(-\frac{5}{2}, 2\right)$
  • C
    $\left(-\frac{3}{2}, \frac{5}{2}\right)$
  • D
    $(-4,0)$
Answer
(D)
Image
Since, the circle touches Y -axis at $(0,2)$.
∴ centre of the circle is $(h, 2)$.
Now, $CA ^2= CB ^2$
$\begin{array}{l}(h-0)^2+(2-2)^2 \\=(h-(-1))^2+(2-0)^2 \\\Rightarrow h^2=h^2+2 h+1+4 \\\Rightarrow 2 h+5=0 \Rightarrow h=-\frac{5}{2}\end{array}$
∴ equation of circle is
$\begin{array}{l}\left(x+\frac{5}{2}\right)^2+(y-2)^2=\left(-\frac{5}{2}\right)^2 \\\Rightarrow x^2+\frac{25}{4}+5 x+y^2-4 y+4=\frac{25}{4} \\\Rightarrow x^2+y^2+5 x-4 y+4=0\end{array}$
Point $(-4,0)$ satisfies this equation.
$\therefore $ option (D) is the correct answer.
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MCQ 1042 Marks
The equation of the circle which touches X-axis at (3, 0) and passes through (1, 4) is given by
  • A
    $x^2+y^2-6 x-5 y+9=0$
  • B
    $x^2+y^2+6 x+5 y-9=0$
  • C
    $x^2+y^2-6 x+5 y-9=0$
  • D
    $x^2+y^2+6 x-5 y+9=0$
Answer
(A)
Image
Since the circle touches X -axis at $(3,0)$.
∴ centre of the circle is $(3, k )$. ${ }^{ }$
Now, $CA ^2= CB ^2$
$\therefore(3-3)^2+( k -0)^2$
$=(3-1)^2+(k-4)^2$
$\therefore \quad k^2=4+k^2-8 k+16$
$\therefore \quad k=\frac{5}{2}$
∴ the required equation of circle is
$(x-3)^2+\left(y-\frac{5}{2}\right)^2=\left(\frac{5}{2}\right)^2$
$\Rightarrow x^2+y^2-6 x-5 y+9=0$
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MCQ 1052 Marks
ABCD is a square, the length of whose side is a. Taking AB and AD as the coordinate axes, the equation of the circle passing through the vertices of the square is
  • A
    $x^2+y^2+a x+a y=0$
  • ✓
    $x^2+y^2- a x- a y=0$
  • C
    $x^2+y^2+2 a x+2 a y=0$
  • D
    $x^2+y^2-2 a x-2 a y=0$
Answer
Correct option: B.
$x^2+y^2- a x- a y=0$
(B)
Image
According to the figure, $A (0,0), B ( a , 0)$, $C ( a , a )$ and $D (0, a )$.
and centre is $\left(\frac{ a }{2}, \frac{ a }{2}\right)$.
∴ the equation of the circle is
$\begin{array}{l}\left(x-\frac{a}{2}\right)^2+\left(y-\frac{a}{2}\right)^2=\frac{a^2}{2} \\\Rightarrow x^2+y^2-a x-a y=0\end{array}$
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MCQ 1062 Marks
The equation of the circle passing through the origin and cutting intercepts of length 3 and 4 units from the positive axes, is
  • A
    $x^2+y^2+6 x+8 y+1=0$
  • B
    $x^2+y^2-6 x-8 y=0$
  • C
    $x^2+y^2+3 x+4 y=0$
  • ✓
    $x^2+y^2-3 x-4 y=0$
Answer
Correct option: D.
$x^2+y^2-3 x-4 y=0$
(D)
Image
Given, $OA =3$ and
OB = 4
$\therefore OL =\frac{3}{2}$ and $CL =2$
By pythagoras theorem,
$OC ^2= OL ^2+ LC ^2$
$OC ^2=\left(\frac{3}{2}\right)^2+2^2$
$=\frac{25}{4}$
$\therefore OC =\frac{5}{2}$
The centre of the circle is $\left(\frac{3}{2}, 2\right)$ and radius $=\frac{5}{2}$.
∴ the equation of the circle is
$\begin{aligned}\left(x-\frac{3}{2}\right)^2+(y-2)^2 & =\left(\frac{5}{2}\right)^2 \\ \therefore \quad x^2+y^2-3 x-4 y & =0\end{aligned}$
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MCQ 1072 Marks
The equation of the circle with centre (2, 2) which passes through (4, 5) is
  • A
    $x^2+y^2-4 x+4 y-77=0$
  • ✓
    $x^2+y^2-4 x-4 y-5=0$
  • C
    $x^2+y^2+2 x+2 y-59=0$
  • D
    $x^2+y^2-2 x-2 y-23=0$
Answer
Correct option: B.
$x^2+y^2-4 x-4 y-5=0$
(B)
Centre $(2,2)$ and
$\begin{aligned} r & =\sqrt{(4-2)^2+(5-2)^2} \\ & =\sqrt{13}\end{aligned}$
Hence, required equation is $\begin{array}{l}(x-2)^2+(y-2)^2=(\sqrt{13})^2 \\\Rightarrow x^2+y^2-4 x-4 y-5=0\end{array}$
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MCQ 1082 Marks
The equation of the circle of radius 5 and touching the coordinate axes in third quadrant is
  • A
    $(x-5)^2+(y+5)^2=25$
  • B
    $(x+4)^2+(y+4)^2=25$
  • C
    $(x+6)^2+(y+6)^2=25$
  • ✓
    $(x+5)^2+(y+5)^2=25$
Answer
Correct option: D.
$(x+5)^2+(y+5)^2=25$
(D)
Since circle touches the co-ordinate axes in III quadrant.
Image
$\therefore $ Radius $=- h =- k$
Hence, $h = k =-5$
∴ Equation of circle is $(x+5)^2+(y+5)^2=25$
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MCQ 1092 Marks
The equation of the circle in the first quadrant which touches each axis at a distance 5 from the origin, is
  • A
    $x^2+y^2+5 x+5 y+25=0$
  • ✓
    $x^2+y^2-10 x-10 y+25=0$
  • C
    $x^2+y^2-5 x-5 y+25=0$
  • D
    $x^2+y^2+10 x+10 y+25=0$
Answer
Correct option: B.
$x^2+y^2-10 x-10 y+25=0$
(B)
The centre of the circle which touches each axis in first quadrant at a distance 5 , will be $(5,5)$ and radius will be 5 .
∴ equation of the circle is
$\begin{array}{l}(x-5)^2+(y-5)^2=(5)^2 \\\Rightarrow x^2+y^2-10 x-10 y+25=0\end{array}$
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MCQ 1102 Marks
The equation of director circle of the circle $x^2+y^2= a ^2$ is
  • A
    $x^2+y^2=4 a^2$
  • ✓
    $x^2+y^2=\sqrt{2} a ^2$
  • C
    $x^2+y^2-2 a ^2=0$
  • D
    None of these
Answer
Correct option: B.
$x^2+y^2=\sqrt{2} a ^2$
(B)
Director circle has its radius $\sqrt{2}$ times that of radius of the given circle.
$\therefore$ The required equation is $x^2+y^2=2 a ^2$.
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MCQ 1112 Marks
The square of the length of the tangent from (3, -4) on the circle $x^2+y^2-4 x-6 y+3=0$ is
  • A
    20
  • B
    30
  • ✓
    40
  • D
    50
Answer
Correct option: C.
40
(C)
Length of tangent
$=\sqrt{3^2+(-4)^2-4(3)-6(-4)+3}=\sqrt{40}$
∴ Square of length of tangent $=40$
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MCQ 1122 Marks
Square of the length of the tangent drawn from the point $(\alpha, \beta)$ to the circle $a x^2+a y^2=r^2$ is
  • A
    $a \alpha^2+a \beta^2-r^2$
  • ✓
    $\alpha^2+\beta^2-\frac{r^2}{a}$
  • C
    $\alpha^2+\beta^2+\frac{ r ^2}{ a }$
  • D
    $\alpha^2+\beta^2-r^2$
Answer
Correct option: B.
$\alpha^2+\beta^2-\frac{r^2}{a}$
(B)
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MCQ 1132 Marks
If the length of the tangent segment from the point (5, 3) to the circle
$x^2+y^2+10 x+ k y-17=0$ is 7, then k equals
  • ✓
    -6
  • B
    4
  • C
    -3
  • D
    10
Answer
Correct option: A.
-6
(A)
Length of tangent segment
$\begin{array}{l}=\sqrt{5^2+3^2+10(5)+k(3)-17}=7 \\\Rightarrow 67+3 k=49 \\\Rightarrow k=-6\end{array}$
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MCQ 1142 Marks
The length of the tangent from the origin to the circle $3 x^2+3 y^2-4 x-6 y+2=0$ is
  • A
    $\sqrt{2}$
  • ✓
    $\frac{\sqrt{2}}{\sqrt{3}}$
  • C
    $\frac{2 \sqrt{2}}{\sqrt{3}}$
  • D
    $\frac{1}{\sqrt{3}}$
Answer
Correct option: B.
$\frac{\sqrt{2}}{\sqrt{3}}$
(B)
Equation of the circle is
$3 x^2+3 y^2-4 x-6 y+2=0$
$\Rightarrow x^2+y^2-\frac{4 x}{3}-\frac{6 y}{3}+\frac{2}{3}=0$
∴ Length of the tangent from the origin is
$\sqrt{0^2+0^2-\left(\frac{4}{3}\right)(0)-\left(\frac{6}{3}\right)(0)+\frac{2}{3}}=\sqrt{\frac{2}{3}}$
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MCQ 1152 Marks
The length of the tangent from the point (-3, 8) to the circle $x^2+y^2-8 x+2 y+1=0$ is
  • A
    $\sqrt{91}$
  • ✓
    $\sqrt{114}$
  • C
    $\sqrt{79}$
  • D
    $\sqrt{131}$
Answer
Correct option: B.
$\sqrt{114}$
(B)
Length of the tangent from the point $(-3,8)$ is
$\sqrt{(-3)^2+8^2-8(-3)+2(8)+1}$
$=\sqrt{9+64+24+16+1}=\sqrt{114}$
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MCQ 1162 Marks
The length of tangent from the point (2, -3) to the circle $2 x^2+2 y^2=1$ is
  • A
    5
  • B
    $10 \sqrt{2}$
  • ✓
    $\frac{5}{\sqrt{2}}$
  • D
    $5 \sqrt{2}$
Answer
Correct option: C.
$\frac{5}{\sqrt{2}}$
(C)
Equation of the circle is $2 x^2+2 y^2-1=0$
$\Rightarrow x^2+y^2-\frac{1}{2}=0$
Length of the tangent from the point $(2,-3)$ is
$\sqrt{2^2+(-3)^2-\frac{1}{2}}=\sqrt{13-\frac{1}{2}}=\frac{5}{\sqrt{2}}$
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MCQ 1172 Marks
The circles $x^2+y^2=9$ and $x^2+y^2-12 y+27= 0$ touch each other. The equation of their common tangent is
  • A
    4y = 9
  • ✓
    y = 3
  • C
    y = -3
  • D
    y = 2
Answer
Correct option: B.
y = 3
(B)
Let $S_1=x^2+y^2-12 y+27=0$
and $S _2=x^2+y^2-9=0$
Then equation of common tangent is
$\begin{array}{l}S_1-S_2=0 \\\Rightarrow-12 y+36=0 \\\Rightarrow y=3\end{array}$
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MCQ 1182 Marks
The two circles $x^2+y^2-2 x+6 y+6=0$ and $x^2+y^2-5 x+6 y+15=0$ touch each other. The equation of their common tangent is
  • ✓
    x = 3
  • B
    y = 6
  • C
    7x - 12y - 21 - 0
  • D
    7x + 12y + 21 = 0
Answer
Correct option: A.
x = 3
(A)
Let $S _1=x^2+y^2-2 x+6 y+6=0$
and $S _2 \equiv x^2+y^2-5 x+6 y+15=0$
Then equation of common tangent is
$\begin{array}{l}S_1-S_2=0 \\\Rightarrow 3 x=9 \\\Rightarrow x=3\end{array}$
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MCQ 1192 Marks
The value of c, for which the line y = 2x + c is a tangent to the circle $x^2+y^2=16$, is
  • A
    $-16 \sqrt{5}$
  • B
    20
  • ✓
    $4 \sqrt{5}$
  • D
    $16 \sqrt{5}$
Answer
Correct option: C.
$4 \sqrt{5}$
(C)
$c= \pm a \sqrt{1+m^2}$
Here, $a=4, m=2$
$\therefore c= \pm 4 \sqrt{1+4}= \pm 4 \sqrt{5}$
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MCQ 1202 Marks
The line $\sqrt{3} x+y- c = 0$ is a tangent to the circle $x^2+y^2=4$, if c is equal to
  • A
    $\pm 16$
  • ✓
    $\pm 4$
  • C
    $\pm 1$
  • D
    $\pm 2$
Answer
Correct option: B.
$\pm 4$
(B)
The line $y=m x+c$ is a tangent to the circle
$x^2+y^2=a^2$, if $c^2=a^2 m^2+a^2$
Here, $a =2, m=-\sqrt{3}$
$\therefore c^2=4(3)+4=16$
$\Rightarrow c= \pm 4$
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MCQ 1212 Marks
If the line x = 7 touches the circle $x^2+y^2-4 x-6 y-12=0$ then the co-ordinates of the point of contact are
  • ✓
    (7, 3)
  • B
    (7, 4)
  • C
    (7, 8)
  • D
    (7, 2)
Answer
Correct option: A.
(7, 3)
(A)
Putting $x=7$, we get $y^2-6 y+9=0$
$\Rightarrow y=3,3$
Hence, the point of contact is $(7,3)$.
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MCQ 1222 Marks
If the line y= mx + c be a tangent to the circle $x^2+y^2=a^2$, then the point of contact is
  • A
    $\left(\frac{- a ^2}{ c }, a ^2\right)$
  • B
    $\left(\frac{ a ^2}{ c }, \frac{- a ^2 m}{ c }\right)$
  • ✓
    $\left(\frac{-a^2 m}{c}, \frac{a^2}{c}\right)$
  • D
    $\left(\frac{-a^2 c}{m}, \frac{a^2}{m}\right)$
Answer
Correct option: C.
$\left(\frac{-a^2 m}{c}, \frac{a^2}{c}\right)$
(C)
Find points of intersection by simultaneously solving for $x$ and $y$ from $y= m x+ c$ and $x^2+y^2=a^2$ which comes out as $\left(-\frac{a^2 m}{c}, \frac{a^2}{c}\right)$
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MCQ 1232 Marks
The equation of tangent to the circle at $x=5 \cos \theta, y=5 \sin \theta, \theta=\frac{\pi}{3}$ is
  • ✓
    $x+y \sqrt{3}=10$
  • B
    $x+y=10$
  • C
    $x \sqrt{3}+y=10$
  • D
    $x \sqrt{3}+y \sqrt{3}=0$
Answer
Correct option: A.
$x+y \sqrt{3}=10$
(A)The equation of the tangent to the circle
$x^2+y^2= a ^2$ at $P (\theta)$ is $x \cos \theta+y \sin \theta= a$
Here, $a=5, \theta=\frac{\pi}{3}$
The equation of the tangent is
$x \cos \frac{\pi}{3}+y \sin \frac{\pi}{3}=5$
$\Rightarrow x\left(\frac{1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=5$
$\Rightarrow x+y \sqrt{3}=10$
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MCQ 1242 Marks
The equation of the tangent to the circle $x^2+y^2-x+3 y=10$ at $(- 2, 1)$ is
  • ✓
    x - y + 3 = 0
  • B
    x - y + 1 = 0
  • C
    2x - y + 3 = 0
  • D
    x + y - 10 =
Answer
Correct option: A.
x - y + 3 = 0
(A)
Equation of tangent is given by
$x x_1+y y_1+ g \left(x+x_1\right)+ f \left(y+y_1\right)+ c =0$
$\therefore-2 x+y-\frac{1}{2}(x-2)+\frac{3}{2}(y+1)-10=0$
$\Rightarrow-4 x+2 y-x+2+3 y+3-20=0$
$\Rightarrow-5 x+5 y-15=0$
$\Rightarrow x-y+3=0$
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MCQ 1252 Marks
The equation of the tangent to the circle $x^2+y^2+2 x-1=0$ at $(-1, \sqrt{2})$ is
  • ✓
    $y-\sqrt{2}=0$
  • B
    $y+x-\sqrt{2}=$
  • C
    $x+\sqrt{2}=0$
  • D
    $x-\sqrt{2}=0$
Answer
Correct option: A.
$y-\sqrt{2}=0$
(A)
The equation of the tangent to the circle
$x^2+y^2+2 g x+2 f y+c=0$ at $\left(x_1, y_1\right)$ is
$xx_1+y y_1+ g \left(x+x_1\right)+ f \left(y+y_1\right)+ c =0$
Here, $g =1, f =0, c =-1$
∴ The equation of the tangent at $(-1, \sqrt{2})$ is $-x+\sqrt{2} y+x-1-1=0$
$\Rightarrow \sqrt{2} y=2 \quad \Rightarrow y=\sqrt{2}$
$\Rightarrow y-\sqrt{2}=0$
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MCQ 1262 Marks
The equation of the tangent to the circle $x^2+y^2= r ^2$ at (a, b) is $a x+b y-\lambda=0$, where $\lambda$ is
  • A
    $a^2$
  • B
    $b^2$
  • ✓
    $r^2$
  • D
    None of these
Answer
Correct option: C.
$r^2$
(C)
Equation of tangent at $( a , b )$ is
$a x+b y-r^2=0$
Comparing with $ax + b y-\lambda=0$, we get $\lambda=r^2$
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MCQ 1272 Marks
The gradient of the tangent at (6, 8) on the circle $x^2+y^2=100$ is
  • A
    $\frac{3}{4}$
  • ✓
    $\frac{-3}{4}$
  • C
    $\frac{4}{3}$
  • D
    $\frac{-3}{2}$
Answer
Correct option: B.
$\frac{-3}{4}$
(B)
Equation of tangent at $(6,8)$ to
$x^2+y^2=100$ is $6 x+8 y=100$
$\therefore y=\frac{-6}{8} x+\frac{100}{8}$
$\Rightarrow m=\frac{-6}{8}=\frac{-3}{4}$
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MCQ 1282 Marks
A circle with centre (a, b) passes through origin. The equation of the tangent to the cit at the origin is
  • A
    ax - by = 0
  • ✓
    ax + by = 0
  • C
    bx - ay = 0
  • D
    bx + ay = 0
Answer
Correct option: B.
ax + by = 0
(B)The slope of the tangent will be
Image
Hence, the equation of the tangent is $y=-\frac{ a }{ b } x$
i.e., $b y+ a x=0$
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MCQ 1292 Marks
The equation of the tangent to the circle $x^2+y^2=17$ at the point (1, -4) is
  • A
    x - 4y = 10
  • ✓
    x - 4y = 17
  • C
    x + 4y = 15
  • D
    x + 4y = - 10
Answer
Correct option: B.
x - 4y = 17
(B)
The equation of the tangent to the circle $x^2+y^2= a ^2$ at $\left(x_1, y_1\right)$ is $x x_1+y y_1= a ^2$
Here, $x_1=1, y_1=-4$
$\therefore$ The equation of the tangent at $(1,-4)$ is
$x-4 y=17$
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MCQ 1302 Marks
The centre of the circle x = 1 + 2 cos $\theta$, $y=3+2 \sin \theta$, is
  • A
    (1, -3)
  • ✓
    (-1, 3)
  • C
    (1, 3)
  • D
    (-1, -3)
Answer
Correct option: B.
(-1, 3)
(B)
$\frac{x+1}{2}=\cos \theta$   ...(i)
and $\frac{y-3}{2}=\sin \theta$   (ii)
Squaring (i) and (ii) and adding, we get
$\left(\frac{x+1}{2}\right)^2+\left(\frac{y-3}{2}\right)^2=1$
$\Rightarrow(x+1)^2+(y-3)^2=4$,
$\therefore$ Centre is $(-1,3)$.
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MCQ 1312 Marks
The parametric representation of the circle $(x-3)^2+(y+4)^2=25$ is
  • A
    $x=5+3 \cos \theta, y=5-3 \sin \theta$
  • B
    $x=5+3 \cos \theta, y=5+3 \sin \theta$
  • ✓
    $x=3+5 \cos \theta, y=-4+5 \sin \theta$
  • D
    $x=3+5 \cos \theta, y=-3+5 \sin \theta$
Answer
Correct option: C.
$x=3+5 \cos \theta, y=-4+5 \sin \theta$
(C)
$(x-3)^2+(y+4)^2=5^2$
Comparing with $(x- h )^2+(y- k )^2= r ^2$, we get
$h=3, k=-4, r=5$
$\therefore $ Parametric equations are $x=3+5 \cos \theta, y=-4+5 \sin \theta$
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MCQ 1322 Marks
The parametric form of the equation of circle $4 x^2+4 y^2=9$ is
  • ✓
    $x=\frac{3}{2} \cos \theta, y=\frac{3}{2} \sin \theta$
  • B
    $x=\frac{2}{5} \sin \theta, y=\frac{2}{5} \cos \theta$
  • C
    $x=\frac{3}{4} \sin \theta, y=\frac{3}{4} \cos \theta$
  • D
    $x=3 \sin \theta, y=\sqrt{2} \cos \theta$
Answer
Correct option: A.
$x=\frac{3}{2} \cos \theta, y=\frac{3}{2} \sin \theta$
(A)
$4 x^2+4 y^2=9$
$\Rightarrow x^2+y^2=\frac{9}{4} \Rightarrow x^2+y^2=\left(\frac{3}{2}\right)^2$
$\therefore x=\frac{3}{2} \cos \theta, y=\frac{3}{2} \sin \theta$
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MCQ 1332 Marks
If the line x + 2by + 7 = 0 is a diameter of the circle $x^2+y^2-6 x+2 y=0$ then b =
  • A
    3
  • B
    -5
  • C
    -1
  • ✓
    5
Answer
Correct option: D.
5
(D)
Here, the centre of circle $(3,-1)$ must lie on the line $x+2 by+7=0$.
$\therefore \ 3-2 b+7=0$
$\Rightarrow b=5$
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MCQ 1342 Marks
Which of the following line is a diameter of the circle $x^2+y^2-6 x-8 y-9=0 ?$
  • A
    3x - 4y = 0
  • B
    4x - y = 0
  • ✓
    x + y = 7
  • D
    x - y = 1
Answer
Correct option: C.
x + y = 7
(C)
Centre $(3,4)$ of the given circle is satisfying only $x+y=7$
∴ Option $( C )$ is the correct answer.
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MCQ 1362 Marks
The circle $x^2+y^2+4 x-4 y+4=0$ touches
  • A
    X-axis
  • B
    Y-axis
  • ✓
    X-axis and Y-axis
  • D
    None of these
Answer
Correct option: C.
X-axis and Y-axis
(C)
Both axis, as centre is $(-2,2)$ and radius is 2 .
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MCQ 1372 Marks
If the radius of the circle $x^2+y^2+2 g x+2 f y+c=0$ is r, then it will touch both the axes, if
  • A
    g = f = c
  • B
    g = f = c = r
  • ✓
    $g=f=\sqrt{c}=r$
  • D
    $g = f$ and $c ^2= r$
Answer
Correct option: C.
$g=f=\sqrt{c}=r$
(C)
Given conditions are $g=f=r$
and $\sqrt{g^2+f^2-c}=r$
$\Rightarrow g=\sqrt{c}=f=r$
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MCQ 1382 Marks
If the circle $x^2+y^2+2 g x+2 f y+c=0$ touches X-axis, then
  • A
    $g=f$
  • ✓
    $g ^2= c$
  • C
    $f ^2= c$
  • D
    $g ^2+ f ^2= c$
Answer
Correct option: B.
$g ^2= c$
(B)
Circle $x^2+y^2+2 g x+2 f y+c=0$ touches X-axis
$\therefore $ radius $=$ ordinate of centre
$\Rightarrow \sqrt{g^2+f^2-c}=(-f)$
$\Rightarrow g ^2= c$
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MCQ 1392 Marks
For the circle $x^2+y^2+6 x-8 y+9=0$,  which of the following statements is true?
  • A
    Circle passes through the point (-3, 4)
  • ✓
    Circle touches X-axis
  • C
    Circle touches Y-axis
  • D
    None of these
Answer
Correct option: B.
Circle touches X-axis
(B)
Intercept made by the circle on the X -axis
$=2 \sqrt{9-9}=0$... [Using Shortcut 2]
$\therefore$ Intercept cut on X -axis is zero.
Hence, circle touches $X$-axis.
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MCQ 1402 Marks
For the circle $x^2+y^2+3 x+3 y=0$, which of the following relation is true?
  • A
    Centre lies on X-axis
  • B
    Centre lies on Y-axis
  • C
    Centre is at origin
  • ✓
    Circle passes through origin
Answer
Correct option: D.
Circle passes through origin
(D)
If $c=0$, circle passes through origin.
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MCQ 1412 Marks
$ax ^2+2 y^2+2 b x y+2 x-y+ c =0$ represents a circle through the origin, if
  • A
    a = 0, b = 0, c = 2
  • B
    a = 1, b = 0, c = 0
  • C
    a = 2,b = 2, c = 0
  • ✓
    a = 2,b = 0, c = 0
Answer
Correct option: D.
a = 2,b = 0, c = 0
(D)
The given equation represents a circle.
if coeff, of $x^2=\operatorname{coeff}$. of $y^2$ and coeff. of $x y=0 $
$\therefore a=2$ and $b=0$
Also, it passes through origin.
$\therefore c=0$
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MCQ 1422 Marks
The equation $a x^2+b y^2+2 h x y+2 g x+2 f y+c=0$ will represent a circle, if
  • A
    a = b = 0 and c = 0
  • B
    f = g and h = 0
  • ✓
    $a=b \neq 0$ and $h=0$
  • D
    f = g and c = 0
Answer
Correct option: C.
$a=b \neq 0$ and $h=0$
(C)
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MCQ 1432 Marks
Radius of the circle $x^2+y^2+2 x \cos \theta+2 y \sin \theta-8=0$ is
  • A
    1
  • ✓
    3
  • C
    $2 \sqrt{3}$
  • D
    $\sqrt{10}$
Answer
Correct option: B.
3
(B)
Radius $=\sqrt{\cos ^2 \theta+\sin ^2 \theta+8}=3$
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MCQ 1442 Marks
The centre and radius of the circle $2 x^2+2 y^2-x=0$ are
  • ✓
    $\left(\frac{1}{4}, 0\right)$ and $\frac{1}{4}$
  • B
    $\left(-\frac{1}{2}, 0\right)$ and $\frac{1}{2}$
  • C
    $\left(\frac{1}{2}, 0\right)$ and $\frac{1}{2}$
  • D
    $\left(0,-\frac{1}{4}\right)$ and $\frac{1}{4}$
Answer
Correct option: A.
$\left(\frac{1}{4}, 0\right)$ and $\frac{1}{4}$
(A)
Here, $g =\frac{-1}{4}, f =0$ and $c =0$
$\therefore$ centre $=(-g,-f)=\left(\frac{1}{4}, 0\right)$
and $r=\sqrt{\frac{1}{16}+0-0}=\frac{1}{4}$
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MCQ 1452 Marks
If the radius of the circle $x^2+y^2-18 x+12 y+ k =0$ is 11, then k =
  • A
    347
  • B
    4
  • ✓
    -4
  • D
    49
Answer
Correct option: C.
-4
(C)
$\begin{array}{l}(\text { Radius })^2=g^2+f^2-c \\ \Rightarrow 121=81+36-k \Rightarrow k=-4\end{array}$
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MCQ 1462 Marks
The equation $x^2+y^2+4 x+6 y+13=0$ represents a
  • A
    circle
  • B
    pair of coincident straight lines
  • C
    pair of concurrent straight lines
  • ✓
    point circle
Answer
Correct option: D.
point circle
(D)
Here, $g =2, f =3$ and $c =13$
$\therefore \quad r=\sqrt{g^2+f^2-c}$
$\therefore \quad r=\sqrt{4+9-13}=0$
option (D) is the correct answer.
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MCQ 1472 Marks
The circle represented by the equation $x^2+y^2+2 g x+2 f y+c=0$ will be a point circle, if
  • ✓
    $g^2+f^2=c$
  • B
    $g ^2+ f ^2> c$
  • C
    $g ^2+ f ^2+ c =0$
  • D
    $g ^2+ f ^2< c$
Answer
Correct option: A.
$g^2+f^2=c$
(A)
Using condition of point circle,
$\begin{array}{l}\text { Radius }=\sqrt{g^2+f^2-c}=0 \\\Rightarrow g^2+f^2-c\end{array}$
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MCQ 1482 Marks
Radius of circle (x - 5)(x - 1) + (y - 7)(y - 4) = 0 is
  • A
    3
  • B
    4
  • ✓
    $\frac{5}{2}$
  • D
    $\frac{7}{2}$
Answer
Correct option: C.
$\frac{5}{2}$
(C)
Extremities of diameter are $(5,7)$ and $(1,4)$.
Radius is half of the distance between them.
$\therefore $ Radius $=\frac{1}{2} \sqrt{(4)^2+(3)^2}$
$=\frac{5}{2}$
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MCQ 1492 Marks
The equation of a circle whose diameter is the line joining the points (-4, 3) and (12, -1) is
  • A
    $x^2+y^2+8 x+2 y+51=0$
  • B
    $x^2+y^2+8 x-2 y-51=0$
  • C
    $x^2+y^2+8 x+2 y-51=0$
  • ✓
    $x^2+y^2-8 x-2 y-51=0$
Answer
Correct option: D.
$x^2+y^2-8 x-2 y-51=0$
(D)
By diameter form, the required equation is
$\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$
$\therefore \quad(x+4)(x-12)+(y-3)(y+1)=0$
$\therefore \quad x^2+y^2-8 x-2 y-51=0$
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MCQ 1502 Marks
The equation $\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$ represents a circle whose centre is
  • A
    $\left(\frac{x_1-x_2}{2}, \frac{y_1-y_2}{2}\right)$
  • ✓
    $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
  • C
    $\left(x_1, y_2\right)$
  • D
    $\left(x_2, y_2\right)$
Answer
Correct option: B.
$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
(B)
The given equation represents a circle having line segment joining $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ as a diameter.
∴ the coordinates of its centre are
$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
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MCQ 1512 Marks
A circle which passes through origin and cuts intercepts on axes a and b, the equation of circle is
  • ✓
    $x^2+y^2-a x-b y=0$
  • B
    $x^2+y^2+a x+b y=0$
  • C
    $x^2+y^2-a x+b y=0$
  • D
    $x^2+y^2+a x-b y=0$
Answer
Correct option: A.
$x^2+y^2-a x-b y=0$
(A)
Centre is $\left(\frac{ a }{2}, \frac{b}{2}\right)$ and radius $=\sqrt{\frac{ a ^2+ b ^2}{4}}$
Image
Hence, equation of circle is
$x^2+y^2-a x-b y=0$
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MCQ 1522 Marks
If $(\alpha, \beta)$ is the centre of a circle passing through the origin, then its equation is
  • A
    $x^2+y^2-\alpha x-\beta y=0$
  • B
    $x^2+y^2+2 \alpha x+2 \beta y=0$
  • ✓
    $x^2+y^2-2 \alpha x-2 \beta y=0$
  • D
    $x^2+y^2+\alpha x+\beta y=0$
Answer
Correct option: C.
$x^2+y^2-2 \alpha x-2 \beta y=0$
(C)
Radius $=$ Distance from origin $=\sqrt{\alpha^2+\beta^2}$
$\therefore(x-\alpha)^2+(y-\beta)^2=\alpha^2+\beta^2$
$\Rightarrow x^2+y^2-2 \alpha x-2 \beta y=0$
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MCQ 1532 Marks
A circle touches the Y-axis at the point (0, 4) and cuts the X-axis in a chord of length 6 units. The radius of the circle is
  • A
    3
  • B
    4
  • ✓
    5
  • D
    6
Answer
Correct option: C.
5
(C)
Let $O^{\prime}$ be the centre
Image
From the figure,
Radius $(r)=\sqrt{(4)^2+(3)^2}=5$
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MCQ 1542 Marks
Equation of circle with centre $(- a,- b)$ and radius $\sqrt{a^2-b^2}$ is
  • ✓
    $x^2+y^2+2 a x+2 b y+2 b^2=0$
  • B
    $x^2+y^2-2 a x-2 b y-2 b^2=0$
  • C
    $x^2+y^2-2 a x-2 b y+2 b^2=0$
  • D
    $x^2+y^2-2 a x+2 b y+2 a ^2=0$
Answer
Correct option: A.
$x^2+y^2+2 a x+2 b y+2 b^2=0$
(A)
$x^2+y^2+2 a x+2 b y+2 b^2=0$
Centre $=(-a,-b)$
∴ option (A) is the correct answer.
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MCQ 1552 Marks
The equation of the circle which touches X-axis and whose centre is (1, 2), is
  • A
    $x^2+y^2-2 x+4 y+1=0$
  • ✓
    $x^2+y^2-2 x-4 y+1=0$
  • C
    $x^2+y^2+2 x+4 y+1=0$
  • D
    $x^2+y^2+4 x+2 y+4=0$
Answer
Correct option: B.
$x^2+y^2-2 x-4 y+1=0$
(B)
Since the circle touches X -axis,
radius $=2$.
$\therefore \quad$ the equation of the circle is
$(x-1)^2+(y-2)^2=2^2$
$\Rightarrow x^2+y^2-2 x-4 y+1=0$
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MCQ 1562 Marks
The equation of the circle which touches both axes and whose centre is (x1, y1) is
  • A
    $x^2+y^2+2 x_1(x+y)+x_1^2=0$
  • ✓
    $x^2+y^2-2 x_1(x+y)+x_1^2=0$
  • C
    $x^2+y^2=x_1^2+y_1^2$
  • D
    $x^2+y^2+2 x x_1+2 y y_1=0$
Answer
Correct option: B.
$x^2+y^2-2 x_1(x+y)+x_1^2=0$
(B)
The equation of circle with centre $\left(x_1, y_1\right)$ is
$\left(x-x_1\right)^2+\left(y-y_1\right)^2=r^2$
Since the circle touches both the axes,
$x_1=y_1= r$
$\therefore\left(x-x_1\right)^2+\left(y-x_1\right)^2=x_1^2$
$\Rightarrow x^2+y^2-2 x_1(x+y)+x_1^2=0$
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MCQ 1572 Marks
The equation of the circle which touches both the axes and whose radius is a, is
  • ✓
    $x^2+y^2-2 a x-2 a y+a^2=0$
  • B
    $x^2+y^2+a x+a y-a^2=0$
  • C
    $x^2+y^2+2 a x+2 a y- a ^2=0$
  • D
    $x^2+y^2- a x- a y+ a ^2=0$
Answer
Correct option: A.
$x^2+y^2-2 a x-2 a y+a^2=0$
(A)
Required equation is $(x-a)^2+(y-a)^2=a^2$
$\Rightarrow x^2+y^2-2 a x-2 a y+a^2=0$
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MCQ 1582 Marks
Centre of the circle $(x-3)^2+(y-4)^2=5$ is
  • ✓
    (3, 4)
  • B
    (-3, -4)
  • C
    (4, 3)
  • D
    (-4, -3)
Answer
Correct option: A.
(3, 4)
(A)
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