Let C_1 be the circle in the third quadrant o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The circle $C_1$ is in the third quadrant with radius $r_1 = 3$ and touches both axes,so its centre is $A(-3, -3)$.The equation of $C_1$ is $(x+3)

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(C) The circle $C_1$ is in the third quadrant with radius $r_1 = 3$ and touches both axes,so its centre is $A(-3, -3)$.The equation of $C_1$ is $(x+3)^2 + (y+3)^2 = 3^2$.The centre of $C_2$ is $B(1, 3)$. Let $r_2$ be the radius of $C_2$.The distance between the centres $A$ and $B$ is $AB = \sqrt{(1 - (-3))^2 + (3 - (-3))^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$.Since the circles touch externally,$AB = r_1 + r_2$. Thus,$2\sqrt{13} = 3 + r_2$,which gives $r_2 = 2\sqrt{13} - 3$.The point of contact $P(\alpha, \beta)$ divides the line segment $AB$ internally in the ratio $r_1 : r_2 = 3 : (2\sqrt{13} - 3)$.Using the section formula:$\alpha = \frac{r_1 x_B + r_2 x_A}{r_1 + r_2} = \frac{3(1) + (2\sqrt{13} - 3)(-3)}{2\sqrt{13}} = \frac{3 - 6\sqrt{13} + 9}{2\sqrt{13}} = \frac{12 - 6\sqrt{13}}{2\sqrt{13}} = \frac{6}{\sqrt{13}} - 3$.$\beta = \frac{r_1 y_B + r_2 y_A}{r_1 + r_2} = \frac{3(3) + (2\sqrt{13} - 3)(-3)}{2\sqrt{13}} = \frac{9 - 6\sqrt{13} + 9}{2\sqrt{13}} = \frac{18 - 6\sqrt{13}}{2\sqrt{13}} = \frac{9}{\sqrt{13}} - 3$.Now,$\beta - \alpha = (\frac{9}{\sqrt{13}} - 3) - (\frac{6}{\sqrt{13}} - 3) = \frac{3}{\sqrt{13}}$.Therefore,$(\beta - \alpha)^2 = (\frac{3}{\sqrt{13}})^2 = \frac{9}{13}$.Here,$m = 9$ and $n = 13$. Since $\operatorname{gcd}(9, 13) = 1$,$m + n = 9 + 13 = 22$.

Let $C_1$ be the circle in the third quadrant of radius $3,$ that touches both coordinate axes. Let $C_2$ be the circle with centre $(1,3)$ that touches $C_1$ externally at the point $(\alpha, \beta)$. If $(\beta-\alpha)^2=\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $m + n$ is equal to :

Similar Questions

If ${c^2} > {a^2}(1 + {m^2})$,then the line $y = mx + c$ will intersect the circle ${x^2} + {y^2} = {a^2}$ at:

Let $x_0, y_0$ be fixed real numbers such that $x_0^2+y_0^2 > 1$. If $x, y$ are arbitrary real numbers such that $x^2+y^2 \leq 1$,then the minimum value of $(x-x_0)^2+(y-y_0)^2$ is

If $(\frac{1}{10}, \frac{-1}{5})$ is the inverse point of a point $(-1, 2)$ with respect to the circle $x^2 + y^2 - 2x + 4y + c = 0$,then $c =$

$A$ line is drawn through a fixed point $P(\alpha, \beta)$ to cut the circle $x^{2}+y^{2}=r^{2}$ at $A$ and $B$. Then $PA \cdot PB$ is equal to

$A$ square is inscribed in the circle $x^2+y^2-10x-6y+30=0$. One side of this square is parallel to $y=x+3$. If $(x_i, y_i)$ are the vertices of the square,then $\sum(x_i^2+y_i^2)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module