MCQ

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
$(x-1)^2+(y+2)^2=5^2$
$\therefore x^2+y^2-2 x+4 y-20=0$

View full question & answer→

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)

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 $=2 k =\frac{10}{3}$

View full question & answer→

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)

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.

View full question & answer→

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)

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
$(x-3)^2+\left(y-\frac{5}{2}\right)^2=\left(\frac{5}{2}\right)^2$
$\Rightarrow x^2+y^2-6 x-5 y+9=0$

View full question & answer→

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)

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
$\begin{array}{l}\left(x-\frac{a}{2}\right)^2+\left(y-\frac{a}{2}\right)^2=\frac{a^2}{2} \\\Rightarrow x^2+y^2-a x-a y=0\end{array}$

View full question & answer→

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)

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
$\begin{aligned}\left(x-\frac{3}{2}\right)^2+(y-2)^2 & =\left(\frac{5}{2}\right)^2 \\ \therefore \quad x^2+y^2-3 x-4 y & =0\end{aligned}$

View full question & answer→

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 $\begin{array}{l}(x-2)^2+(y-2)^2=(\sqrt{13})^2 \\\Rightarrow x^2+y^2-4 x-4 y-5=0\end{array}$

View full question & answer→

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.

$\therefore $ Radius $=- h =- k$
Hence, $h = k =-5$
∴ Equation of circle is $(x+5)^2+(y+5)^2=25$

View full question & answer→

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
$\begin{array}{l}(x-5)^2+(y-5)^2=(5)^2 \\\Rightarrow x^2+y^2-10 x-10 y+25=0\end{array}$

View full question & answer→

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 $x^2+y^2=2 a ^2$.

View full question & answer→

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.

View full question & answer→

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)

View full question & answer→

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
$\begin{array}{l}=\sqrt{5^2+3^2+10(5)+k(3)-17}=7 \\\Rightarrow 67+3 k=49 \\\Rightarrow k=-6\end{array}$

View full question & answer→

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
$\sqrt{0^2+0^2-\left(\frac{4}{3}\right)(0)-\left(\frac{6}{3}\right)(0)+\frac{2}{3}}=\sqrt{\frac{2}{3}}$

View full question & answer→

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
$\sqrt{(-3)^2+8^2-8(-3)+2(8)+1}$
$=\sqrt{9+64+24+16+1}=\sqrt{114}$

View full question & answer→

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
$\sqrt{2^2+(-3)^2-\frac{1}{2}}=\sqrt{13-\frac{1}{2}}=\frac{5}{\sqrt{2}}$

View full question & answer→

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
$\begin{array}{l}S_1-S_2=0 \\\Rightarrow-12 y+36=0 \\\Rightarrow y=3\end{array}$

View full question & answer→

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
$\begin{array}{l}S_1-S_2=0 \\\Rightarrow 3 x=9 \\\Rightarrow x=3\end{array}$

View full question & answer→

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.

$4 \sqrt{5}$

View full question & answer→

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, $a =2, m=-\sqrt{3}$
$\therefore c^2=4(3)+4=16$
$\Rightarrow c= \pm 4$

View full question & answer→

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 $(7,3)$.

View full question & answer→

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 $\left(-\frac{a^2 m}{c}, \frac{a^2}{c}\right)$

View full question & answer→

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
$x \cos \frac{\pi}{3}+y \sin \frac{\pi}{3}=5$
$\Rightarrow x\left(\frac{1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=5$
$\Rightarrow x+y \sqrt{3}=10$

View full question & answer→

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
$x x_1+y y_1+ g \left(x+x_1\right)+ f \left(y+y_1\right)+ c =0$
$\therefore-2 x+y-\frac{1}{2}(x-2)+\frac{3}{2}(y+1)-10=0$
$\Rightarrow-4 x+2 y-x+2+3 y+3-20=0$
$\Rightarrow-5 x+5 y-15=0$
$\Rightarrow x-y+3=0$

View full question & answer→

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 $-x+\sqrt{2} y+x-1-1=0$
$\Rightarrow \sqrt{2} y=2 \quad \Rightarrow y=\sqrt{2}$
$\Rightarrow y-\sqrt{2}=0$

View full question & answer→

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.

$r^2$

View full question & answer→

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 $6 x+8 y=100$
$\therefore y=\frac{-6}{8} x+\frac{100}{8}$
$\Rightarrow m=\frac{-6}{8}=\frac{-3}{4}$

View full question & answer→

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

Hence, the equation of the tangent is $y=-\frac{ a }{ b } x$
i.e., $b y+ a x=0$

View full question & answer→

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
$x-4 y=17$

View full question & answer→

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 $(-1,3)$.

View full question & answer→

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 $x=3+5 \cos \theta, y=-4+5 \sin \theta$

View full question & answer→

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)
$4 x^2+4 y^2=9$
$\Rightarrow x^2+y^2=\frac{9}{4} \Rightarrow x^2+y^2=\left(\frac{3}{2}\right)^2$
$\therefore x=\frac{3}{2} \cos \theta, y=\frac{3}{2} \sin \theta$

View full question & answer→

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.

(D)
Here, the centre of circle $(3,-1)$ must lie on the line $x+2 by+7=0$.
$\therefore \ 3-2 b+7=0$
$\Rightarrow b=5$

View full question & answer→

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

View full question & answer→

MCQ 1352 Marks

Circle $x^2+y^2+6 y=0$ touches

A
Y-axis at the origin
B
X-axis at the origin
C
X-axis at the point (3, 0)
D
Y-axis at the point (0, 2)

Answer

(B)
Centre is $(0,-3)$ and radius $=\sqrt{0^2+9-0}=3$

Hence, circle touches X-axis at the origin.

View full question & answer→

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

View full question & answer→

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.

$g=f=\sqrt{c}=r$

View full question & answer→

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
$\Rightarrow \sqrt{g^2+f^2-c}=(-f)$
$\Rightarrow g ^2= c$

View full question & answer→

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.

View full question & answer→

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.

View full question & answer→

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.
$\therefore c=0$

View full question & answer→

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)

View full question & answer→

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.

(B)
Radius $=\sqrt{\cos ^2 \theta+\sin ^2 \theta+8}=3$

View full question & answer→

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 $r=\sqrt{\frac{1}{16}+0-0}=\frac{1}{4}$

View full question & answer→

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

View full question & answer→

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.

View full question & answer→

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,
$\begin{array}{l}\text { Radius }=\sqrt{g^2+f^2-c}=0 \\\Rightarrow g^2+f^2-c\end{array}$

View full question & answer→

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 $=\frac{1}{2} \sqrt{(4)^2+(3)^2}$
$=\frac{5}{2}$

View full question & answer→

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
$\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$
$\therefore \quad(x+4)(x-12)+(y-3)(y+1)=0$
$\therefore \quad x^2+y^2-8 x-2 y-51=0$

View full question & answer→

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
$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.

View full question & answer→

Maths MCQ