MCQ 11 Mark
The equation of the incircle formed by the coordinate axes and the line 4x + 3y = 6 is:
- A
$x^2 + y^2 - 6x - 6y + 9 = 0$
- ✓
$4 (x^2 + y^2 - x - y) + 1 = 0$
- C
$4 (x^2 + y^2 + x + y) + 1 = 0$
- D
AnswerCorrect option: B. $4 (x^2 + y^2 - x - y) + 1 = 0$
- $4 (x^2 + y^2 - x - y) + 1 = 0$
Solution:
The line 4x + 3y = 6 cuts the coordinate axes at $\Big(\frac{3}{2},\ 0\Big)$ and (0, 2)
The coordinates of the incentre is $\Big(\frac{\text{ax}_1+\text{bx}_2+\text{cx}_3}{\text{a+b+c}},\ \frac{\text{ay}_1+\text{by}_2+\text{cy}}{\text{a+b+c}}\Big)$
Here, $\text{a}=\frac{5}{2},\ \text{b}=\frac{3}{2},\ \text{c}=2,\ \text{x}_1=0,\ \text{y}_1\\=0,\ \text{x}_2=0,\ \text{y}_2=2,\ \text{x}_3=\frac{3}{2},\ \text{y}_3=0$
Thus, the coordinates of the incentre:
$\Big(\frac{0+0+3}{6},\ \frac{0+3+0}{6}\Big)$
$=\big(\frac{1}{2},\ \frac{1}{2}\Big)$
The equation of the incircle:
$\Big(\text{x}-\frac{1}{2}\Big)^2+\Big(\text{y}-\frac{1}{2}\Big)^2=\text{a}^2$
Also, radius of the incircle $=\frac{\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}}{\text{s}}$
Here, $\text{s}=\frac{\text{a+b+c}}{2}=\frac{\frac{5}{2}+\frac{3}{2}+2}{2}=3$
$\therefore$ Radius of the incircle $=\sqrt{\frac{3(3-\text{a})(3-\text{b}(3-\text{c}))}{3}}$
$=\frac{\sqrt{3\Big(3-\frac{5}{2}\Big)\Big(3-\frac{3}{2}\Big)(3-\text{c})}}{3}$
$=\frac{\sqrt{3\Big(3-\frac{1}{2}\Big)\Big(\frac{3}{2}\Big)}}{3}$
$=\frac{1}{2}$
The equation of circle:
$\Big(\text{x}-\frac{1}{2}\Big)^2+\Big(\text{y}-\frac{1}{2}\Big)^2=\frac{1}{4}$
$\Rightarrow4(\text{x}^2+\text{y}^2)-\text{x}-\text{y}+1=0$ View full question & answer→MCQ 21 Mark
If the centroid of an equilateral triangle is $(1,1)$ and its one vertex is $(-1,2)$, then the equation of its circumcircle is:
- ✓
$x^2+y^2-2 x-2 y-3=0$
- B
$x^2+y^2+2 x-2 y-3=0$
- C
$x^2+y^2+2 x+2 y-3=0$
- D
AnswerCorrect option: A. $x^2+y^2-2 x-2 y-3=0$
- $x^2+y^2-2 x-2 y-3=0$
Solution:
The centre of the circumcircle is (1, 1).
Radius of the circumcircle
$\therefore$ Equation of the circle: $=\sqrt{(1+1)^2+(1-2)^2}=\sqrt{5}$
$(x - 1)^2 + (y - 1)^2 = 5$
$\Rightarrow x^2 + y2 - 2x - 2y - 3 = 0$ View full question & answer→MCQ 31 Mark
The equation of the circle which touches the axes of coordinates and the line $\frac{\text{x}}{3}+\frac{\text{y}}{4}=1$ and whose centres lie in the first quadrant is $x^2 + y^2 − 2cx − 2cy + c^2 = 0$, where c is equal to:
Answer
- 6
Solution:
The equation of the circle that touches the axes of coordinates is $x^2 + y^2 - 2cx − 2cy + c^2 = 0$.
Also, $x^2 + y^2 − 2cx − 2cy + c^2 = 0$ touches the line $\frac{\text{x}}{3}+\frac{\text{y}}{4}=1$ or $4x +3y -12 = 0$.
Since the circle lies in the first quadrant, it centre is is (c, c).
From the figure, we have:
$\Bigg|\frac{4\text{c}+3\text{c}-12}{\sqrt{4^2+3^3}}\Bigg|=\text{c}$
$\Rightarrow\frac{7\text{c}-12}{5}=\text{c}$
$\Rightarrow\text{c}=6$ View full question & answer→MCQ 41 Mark
The equation of the circle passing through the point $(1,1)$ and having two diameters along the pair of lines $x^2-y^2-$ $2 x+4 y-3=0$, is:
- ✓
$x^2+y^2-2 x-4 y+4=0$
- B
$x^2+y^2+2 x+4 y-4=0$
- C
$x^2+y^2-2 x+4 y+4=0$
- D
AnswerCorrect option: A. $x^2+y^2-2 x-4 y+4=0$
- $x^2+y^2-2 x-4 y+4=0$
Solution:
Let the required equation of the circle be $(x - h)^2 + (y - k)^2 = a^2$.
Comparing the given equation $x^2 - y^2 - 2x + 4y - 3 = 0$ with
$ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$, we get:
$a = 1, b = -1, h = 0, g = -1, f = 2, c = -3$
Intersection point $\Big(\frac{\text{hf}-\text{bg}}{\text{ab}-\text{h}^2},\ \frac{\text{gh}-\text{af}}{\text{ab}-\text{h}^2}\Big)=\Big(\frac{-1}{-1},\ \frac{-2}{-1}\Big)=(1,\ 2)$
Thus, the centre of the circle is $(1, 2)$
The equation of the required circle is $(x - 1)^2 + (y - 2)^2 = a^2$
Since circle passes through $(1, 1)$, we have:
$1 = a^2$
$\therefore$ Equation of the required circle:
$(x - 1)^2 + (y - 2)^2 = 1$
$\Rightarrow x^2 + y^2 - 2x - 4y + 4 = 0$ View full question & answer→MCQ 51 Mark
If the point (2, k) lies outside the circles $x^2 + y^2 + x - 2y - 14 = 0$ and $x^2 + y^2 = 13$ then klies in the interval:
- A
$(-3,\ -2)\cup(3,\ 4)$
- B
$-3,\ 4$
- ✓
$(-\infty,\ -3)\cup(4,\ \infty)$
- D
$(-\infty,\ -2)\cup(3,\ \infty)$
AnswerCorrect option: C. $(-\infty,\ -3)\cup(4,\ \infty)$
- $(-\infty,\ -3)\cup(4,\ \infty)$
Solution:
The given equations of the circles are $x^2+y^2+x-2 y-14=0$ and $x^2+y^2=13$.
Since $(2, k)$ lies outside the given circles, we have:
$4+k^2+2-2 k-14>0 \text { and } 4+k^2>13$
$\Rightarrow k^2-2 k-8>0 \text { and } k^2>9$
$\Rightarrow(k-4)(k+2)>0 \text { and } k^2>9$
$\Rightarrow k>4 \text { or } k<-2 \text { and } k>3 \text { or } k<-3$
$\Rightarrow k>4 \text { and } k<-3$
$\Rightarrow k \in(-\infty,-3) \cup(4, \infty)$ View full question & answer→MCQ 61 Mark
The circle $x^2+y^2+2 g x+2 f y+c=0$ does not intersect $x$-axis, if:
AnswerCorrect option: A. $\mathrm{g}^2<\mathrm{c}$
- $\mathrm{g}^2<\mathrm{c}$
Solution:
Given:
$x^2 + y^2 + 2gx + 2fy + c = 0$ ......... (1)
The given circle intersects the x-axis.
The equation of circle becomes $x^2 + 2gx + c = 0$ ......... (2)
Solving equation (2):
$\therefore$ Discriminant, $\text{D}=\sqrt{4\text{g}^2-4\text{c}}\geq0$
$\Rightarrow4\text{g}^2-4\text{c}\geq0$
$\Rightarrow\text{g}^2-\text{c}\geq0$
$\Rightarrow\text{g}^2\geq\text{c}$
Hence, if $g^2 < c$, then the given circle will not intersect the x-axis. View full question & answer→MCQ 71 Mark
The equation $x^2 + y^2 + 2x - 4y + 5 = 0$ represents:
- ✓
- B
A pair of straight lines.
- C
A circle of non-zero radius..
- D
Answer
- A point
Solution:
The radius of the given circle $=\sqrt{1^1+(-2)^2-5}=0$
Hence, the radius of the given circle is zero, which represents a point. View full question & answer→MCQ 81 Mark
If the circles $x^2+y^2=9$ and $x^2+y^2+8 y+c=0$ touch each other, then c is equal to:
Answer
- 15
Solution:
The centre of the circle $x^2+y^2=9$ is $(0,0)$.
Let us denote it by $\mathrm{C}_1$.
The centre of the circle $x^2+y^2+8 y+c=0$ is $(0,-4)$.
Let us denote it by $\mathrm{C}_2$.
The radius of $x^2+y^2=9$ is 3 units.
$x^2+y^2+8 y+c=0$
$\Rightarrow(\text{x}-0)^2+(\text{y}+4)^2=16-\text{c}=(\sqrt{16-\text{c}})^2$
Therefore, the radius of the above circle is $\sqrt{16-\text{c}}$
Let the circles touch each other at P.
$\therefore\text{C}_1\text{C}_2=\text{PC}_2+\text{PC}_1$
$\Rightarrow\text{PC}_2=4-3=1$
$\Rightarrow\text{PC}_2-1=\sqrt{16-\text{c}}$
$\Rightarrow\text{c}=15$ View full question & answer→MCQ 91 Mark
Equation of the circle through origin which cuts intercepts of length a and b on axes is:
- A
$x^2+y^2+a x+b y=0$
- ✓
$x^2+y^2-a x-b y=0$
- C
$x^2+y^2+b x+a y=0$
- D
AnswerCorrect option: B. $x^2+y^2-a x-b y=0$
- $x^2+y^2-a x-b y=0$
Solution:
Centre of the circle is $\Big(\frac{\text{a}}{2},\ \frac{\text{b}}{2}\Big)$ and its radius is $\sqrt{\Big(\frac{\text{a}}{2}\Big)^2+\Big(\frac{\text{b}}{2}\Big)^2}=\frac{1}{2}\sqrt{\text{a}^2+\text{b}^2}$
Equation of circle:
$\Big(\text{x}-\frac{\text{a}}{2}\Big)^2+\Big(\text{y}-\frac{\text{b}}{2}\Big)^2=\frac{1}{4}(\text{a}^2+\text{b}^2)$
$\Rightarrow(2\text{x}-\text{a}^2)+(2\text{y}-\text{b})^2=(\text{a}^2+\text{b}^2)$
$\Rightarrow4\text{x}^2+\text{a}^2-4\text{ax}+4\text{y}^2+\text{b}^2-4\text{by}=\text{a}^2+\text{b}^2$
$\Rightarrow\text{x}^2-\text{ax}+\text{y}^2-\text{by}=0$ View full question & answer→MCQ 101 Mark
The number of integral values of $\lambda$ for which the equation $\text{x}^2+\text{y}^2+\lambda+(1-\lambda)\text{y}+5=0$ is the equation of a circle whose radius cannot exceed 5, is:
Answer$\sqrt{\Big(\frac{-\lambda}{2}^2\Big)+\Big(\frac{\lambda-1}{2}\Big)^2-5}\leq5$
$\Rightarrow\Big(\frac{-\lambda}{2}^2\Big)+\Big(\frac{\lambda-1}{2}\Big)\leq30$
$\lambda^2+(\lambda-1)^2\leq120$
$\Rightarrow2\lambda^2-2\lambda-199\leq0$
Using quadratic formula:
$\Rightarrow\lambda=\frac{2\pm\sqrt{2^2-4(2)(-119)}}{2(2)}$
$\Rightarrow\lambda=\frac{2\pm\sqrt{956}}{4}$
$\Rightarrow\lambda=\frac{1\pm\sqrt{239}}{2}$
$\Rightarrow\lambda=-7.23,\ 8.23$
$\Rightarrow-7.23\leq\lambda\leq8.23$
$\Rightarrow\lambda=-7,\ -6,\ -5,\ -4,\ -3,\ -2,\ -1\\0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ $ $(\text{if}\ \lambda\in\text{Z})$
Thus, the number of integral values of $\lambda$ is 16.
View full question & answer→MCQ 111 Mark
Equation of the diameter of the circle $x^2 + y^2 − 2x + 4y = 0$ which passes through the origin is:
Answer
- 2x + y = 0
Solution:
Let the diameter of the circle be y = mx.
Since the diameter of the circle passes through its centre, (1, -2) satisfies the equation of the diameter.
$\therefore$ m = -2
Substituting the value of m in the equation of diameter:
y = -2x
⇒ 2x + y = 0
Hence, the required equation of the diameter is 2x + y = 0. View full question & answer→MCQ 121 Mark
If $2\text{x}^2+\lambda\text{xy}+2\text{y}^2(\lambda-4)\text{x}+6\text{y}-5=0$ is the equation of a circle, then its radius is:
- A
$3\sqrt{2}$
- B
$2\sqrt{3}$
- C
$2\sqrt{2}$
- ✓
AnswerThe given equation is $2\text{x}^2+\lambda\text{xy}+2\text{y}^2+(\lambda-4)\text{x}+6\text{y}-5=0$ which can be rewritten as
$\text{x}^2+\frac{\lambda\text{xy}}{2}+\text{y}^2+\frac{(\lambda-4)}{2}\text{x}+3\text{y}-\frac{5}{2}=0.$
Comparing the given equation $\text{x}^2+\text{y}62+2\text{gx}+2\text{fy}+\text{c}=0$ with we get: $\lambda=0$
$\therefore\text{x}^2+\text{y}^2-2\text{x}+3\text{y}-\frac{5}{2}=0$
$\therefore$ Radius $=\sqrt{(-1)^2+\Big(\frac{3}{2}\Big)^2+\frac{5}{2}}=\sqrt{1+\frac{9}{4}+\frac{5}{2}}=\sqrt{\frac{23}{4}}=\frac{\sqrt{23}}{2}$
View full question & answer→MCQ 131 Mark
If the circle $x^2 + y^2 + 2ax + 8y + 16 = 0$ touches x-axis, then the value of a is:
- A
$\pm16$
- ✓
$\pm4$
- C
$\pm8$
- D
$\pm1$
AnswerCorrect option: B. $\pm4$
- $\pm4$
Solution:
The equation of the circle is $x^2 + y^2 + 2ax + 8y + 16 = 0$.
Its centre is (-a, -4) and its radius is a units.
Since the circle touches the x-axis, we have:
$\sqrt{(-\text{a}+\text{a})^2+(4-0)^2}=\text{a}$
$\Rightarrow\text{a}=\pm4$ View full question & answer→MCQ 141 Mark
The radius of the circle represented by the equation $3\text{x}^2+3\text{y}^2+(\lambda-6)\text{y}+3=0$ is:
- ✓
$\frac{3}{2}$
- B
$\frac{\sqrt{17}}{2}$
- C
$\frac{2}{3}$
- D
AnswerCorrect option: A. $\frac{3}{2}$
The equation of the circle is $3\text{x}^2+3\text{y}^2+(\lambda-6)\text{y}+3=0$
$\therefore$ Coefficient of $\text{xy}=0$
$\Rightarrow\lambda=0$
$\therefore3\text{x}^2+3\text{y}^2+9\text{x}-6\text{y}+3=0$
$\Rightarrow\text{x}^2+\text{y}^2+3\text{x}-2\text{y}+1=0$
Therefore, the radius of the circle is $\sqrt{\Big(\frac{3}{2}\Big)^2+(-1)^2-1}=\frac{3}{2}.$
View full question & answer→MCQ 151 Mark
The equation of a circle with radius $5$ and touching both the coordinate axes is:
- A
$x^2+y^2 \pm 10 x \pm 10 y+5=0$
- B
$x^2+y^2 \pm 10 x \pm 10 y=0$
- ✓
$x^2+y^2 \pm 10 x \pm 10 y+25=0$
- D
$x^2+y^2 \pm 10 x \pm 10 y+51=0$
AnswerCorrect option: C. $x^2+y^2 \pm 10 x \pm 10 y+25=0$
- $x^2+y^2 \pm 10 x \pm 10 y+25=0$
Solution:
Case I: If the circle lies in the first quadrant:
The equation of a circle that touches both the coordinate axes and hasradius $a$ is $x^2+y^2-2 a x-2 a y+a^2=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2-10 x-10 y+25=0$.
Case II: If the circle lies in the second quadrant:
The equation of a circle that touches both the coordinate axes and has radius a is $x^2+y^2+2 a x-2 a y+a^2=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2+10 x-10 y+25=0$.
Case III: If the circle lies in the third quadrant:
The equation of a circle that touches both the coordinate axes and has radius $a$ is $x^2+y^2+2 a x+2 a y+a^2=0$
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2+10 x+10 y+25=0$.
Case IV: If the circle lies in the fourth quadrant:
The equation of a circle that touches both the coordinate axes and has radius $a$ is $x^2+y^2-2 a x+2 a y+a^2=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2-10 x+10 y+25=0$.
Hence, the required equation of the circle is $x^2+y^2 \pm 10 x \pm 10 y+25=0$. View full question & answer→MCQ 161 Mark
If the point $(\lambda,\ \lambda+1)$ lies inside the region bounded by the curve $\text{x}=\sqrt{25-\text{y}^2}$ and y-axis, then $\lambda$ belongs to the interval:
AnswerCorrect option: A. $(-1,\ 3)$
- $(-1,\ 3)$
Solution:
The given equation of the curve is $x^2 + y^2 = 25$
Since $(\lambda,\ \lambda+1)$ lies inside the region bounded by the curve $x^2 + y^2 = 25$ and the y-axis, we have:
$\lambda^2+(\lambda+1)^2<25,$ provided $\lambda+1>0$
$\Rightarrow\lambda^2+\lambda^2+12\lambda<25,\ \lambda>-1$
$\Rightarrow2\lambda^2+2\lambda-24<0,\ \lambda>-1$
$\Rightarrow\lambda^2+\lambda-12<0,\ \lambda>-1$
$\Rightarrow(\lambda-3)(\lambda+4)<0,\ \lambda>-1$
$\Rightarrow-4<\lambda<3,\ \lambda>-1$
$\Rightarrow\lambda\in(-1,\ 3)$ View full question & answer→MCQ 171 Mark
The area of an equilateral triangle inscribed in the circle $x^2 + y^2 - 6x - 8y - 25 = 0$ is:
- ✓
$\frac{225\sqrt{3}}{6}$
- B
$25\pi$
- C
$50\pi-100$
- D
AnswerCorrect option: A. $\frac{225\sqrt{3}}{6}$
- $\frac{225\sqrt{3}}{6}$
Solution:
Let ABC be the required equilateral triangle.
The equation of the circle is $x^2 + y^2 - 6x - 8y - 25 = 0$.
Therefore, coordinates of the centre O is (3, 4).
Radius of the circle $=\text{OA}=\text{OB}=\text{OC}=\sqrt{9+16+25}=5\sqrt{2}$
In $\Delta\text{BOD},$ we have:
$\sin60^\circ=\frac{\text{DB}}{\text{BO}}$
$\Rightarrow\text{DB}=\frac{\sqrt{3}}{2}(5\sqrt{2})$
$\Rightarrow\text{BC}=2\text{BD}-\sqrt{3}\big(5\sqrt{2}\big)=5\sqrt{6}$
Now, area of $\triangle\text{ABC}=\frac{\sqrt{3}}{4}\text{BC}^2=\big(5\sqrt{6}\big)^2\\=\frac{\sqrt{3}(150)}{4}=\frac{\sqrt{3}(75)}{2}=\frac{\sqrt{3}(225)}{6}$ square units View full question & answer→MCQ 181 Mark
The equation of the circle concentric with $x^2+y^2-3 x+4 y-c=0$ and passing through $(-1,-2)$ is:
- A
$x^2+y^2-3 x+4 y-1=0$
- ✓
$x^2+y^2-3 x+4 y=0$
- C
$x^2+y^2-3 x+4 y+2=0$
- D
AnswerCorrect option: B. $x^2+y^2-3 x+4 y=0$
- $x^2+y^2-3 x+4 y=0$
Solution:
The centre of the circle $x^2+y^2-3 x+4 y - c=0$ is $\Big(\frac{3}{2},\ -2\Big).$
Therefore, the centre of the required circle is $\Big(\frac{3}{2},\ -2\Big).$
The equation of the circle is $\Big(\text{x}-\frac{3}{2}\Big)^2+(\text{y}+2)^2=\text{a}^2. \ ......(1)$
Also, circle (1) passes through (-1, -2).
$\therefore\Big(-1-\frac{3}{2}\Big)^2+\Big(-2+2\Big)^2=\text{a}^2$
$\Rightarrow\text{a}=\frac{5}{2}$
Substituting the value of a in equation (1):
$\Big(\text{x}-\frac{3}{2}\Big)^2+(\text{y}+2)^2=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\frac{(2\text{x}-3)^2}{4}+(\text{y}+2)^2=\frac{25}{4}$
$\Rightarrow(2\text{x}-3)^2+4(\text{y}+2)^2=25$
$\Rightarrow\text{x}^2+\text{y}^2-3\text{x}+4\text{y}=0$
Hence, the required equation of the circle is $x^2+y^2-3 x+4 y=0$. View full question & answer→MCQ 191 Mark
If $(-3,2)$ lies on the circle $x^2+y^2+2 g x+2 f y+c=0$ which is concentric with the circle $x^2+y^2+6 x+8 y-5=0$, then $\mathrm{c}=$
Answer
- -11
Solution:
The centre of the circle $x^2+y^2+6 x+8 y-5=0$ is $(-3,-4)$.
The circle $x^2+y^2+2 g x+2 f y+c=0$ is concentric with the circle $x^2+y^2+6 x+8 y-5=0$.
Thus, the centre of $x^2+y^2+2 g x+2 f y+c=0$ is $(-3,-4)$.
$\therefore g=3, f=4$
Also, it is given that $(-3,2)$ lies on the circle $x^2+y^2+2 g x+2 f y+c=0$.
$\therefore(-3)^2+2^2+2(3)(-3)+2(4)(2)+c=0$
$\Rightarrow 9+4-18+16+c=0$
$\Rightarrow c=-11$ View full question & answer→MCQ 201 Mark
If the equation of a circle is $\lambda\text{x}^2+(2\lambda-3)\text{y}^2-4\text{x}+6\text{y}-1=0,$ then the coordinates of centre are:
- A
$\Big(\frac{4}{3},\ -1\Big)$
- ✓
$\Big(\frac{2}{3},\ -1\Big)$
- C
$\Big(\frac{-2}{3},\ 1\Big)$
- D
$\Big(\frac{2}{3},\ 1\Big)$
AnswerCorrect option: B. $\Big(\frac{2}{3},\ -1\Big)$
- $\Big(\frac{2}{3},\ -1\Big)$
Solution:
To find the centre:
Coefficient of $x^2$ = Coefficient of $y^2$
$\therefore\lambda=2\lambda-3\Rightarrow\lambda=3$
Therefore, the given equation can be rewritten as $3\text{x}^2+3\text{y}^2-4\text{x}+6\text{y}-1=0.$
$\therefore\text{x}^2+\text{y}^2-\frac{4}{3}\text{x}+2\text{y}-\frac{1}{3}=0$
Thus, the coordinates of the centre is $\Big(\frac{2}{3},\ -1\Big).$ View full question & answer→MCQ 211 Mark
If the circles $x^2 + y^2 = a$ and $x^2 + y^2 - 6x - 8y + 9 = 0$, touch externally, then a =
Answer
- 1
Solution:
$x^2+y^2=a$........ (1)
And, $x^2+y^2-6 x-8 y+9=0$ ........ (2)
Let circles (1) and (2) touch each other at point $P$.
The centre of the circle $x^2+y^2=a, 0$, is $(0,0)$.
The centre of the circle $x^2+y^2-6 x-8 y+9=0, C_1$, is $(3,4)$.
Also, radius of circle (1) $=\sqrt{\text{a}}=\text{OP}$
Radius of circle (2) $\sqrt{9+16-9}=4=\text{C}_1\text{P}$
From figure, we have:
$\Rightarrow\sqrt{3^2+4^2}=4+\sqrt{\text{a}}$
$\Rightarrow5=4+\sqrt{\text{a}}$
$\Rightarrow\text{a}=1$ View full question & answer→MCQ 221 Mark
The equation of the circle passing through the origin which cuts off intercept of length 6 and 8 from the axes is:
- A
$x^2+y^2-12 x-16 y=0$
- B
$x^2+y^2+12 x+16 y=0$
- C
$x^2+y^2+6 x+8 y=0$
- ✓
$x^2+y^2-6 x-8 y=0$
AnswerCorrect option: D. $x^2+y^2-6 x-8 y=0$
- $x^2+y^2-6 x-8 y=0$
Solution:
The centre of the required circle is$\Big(\frac{6}{2},\ \frac{8}{2}\Big)=(3,\ 4).$
The radius of the required circle is $\sqrt{3^2+4^2}=\sqrt{25}=5$
Hence, the equation of the circle is as follows:
$(x - 3)^2 + (y - 4)^2 = 52$
$\Rightarrow x^2 + y^2 - 6x - 8y = 0$ View full question & answer→MCQ 231 Mark
If (x, 3) and (3, 5) are the extremities of a diameter of a circle with centre at (2, y), then the values of x and y are:
AnswerThe end points of the diameter of a circle are (x, 3) and (3, 5).
According to the question, we have:
$\frac{\text{x}+3}{2}=2,\ \text{y}=\frac{5+3}{2}$
$\Rightarrow\text{x}=1,\ \text{y}=4$
View full question & answer→MCQ 241 Mark
If the equation $(4a - 3) x^2 + ay^2 + 6x - 2y + 2 = 0$ represents a circle, then its centre is:
Answer
- (-3, 1)
Solution:
If the equation $(4 a-3) x^2+a y^2+6 x-2 y+2=0$ represents a circle, then we have:
Coefficient of $x^2=$ Coefficient of $y^2$
$\Rightarrow 4 a-3=a$
$\Rightarrow a=1$
$\therefore$ Equation of the circle $=x^2+y^2+6 x-2 y+2=0$
Thus, the coordinates of the centre is $(-3,1)$. View full question & answer→