If a circle C passing through (4, 0) touches | vedclass

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(A) Let the given circle be $S_1: x^2 + y^2 + 4x - 6y - 12 = 0$. The center $A$ is $(-2, 3)$ and the radius $r_1 = \sqrt{(-2)^2 + 3^2 - (-12)} = \sqrt

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$5$

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(A) Let the given circle be $S_1: x^2 + y^2 + 4x - 6y - 12 = 0$. The center $A$ is $(-2, 3)$ and the radius $r_1 = \sqrt{(-2)^2 + 3^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5$.Let the center of circle $C$ be $B(h, k)$ and its radius be $r_2$.The point of contact is $O(1, -1)$. Since the circles touch externally,the centers $A, O, B$ are collinear and $O$ divides $AB$ in the ratio $r_1 : r_2$.Using the section formula for $O(1, -1)$ dividing $AB$ internally in ratio $5 : r_2$:$1 = \frac{5h + r_2(-2)}{5 + r_2}$ $\Rightarrow 5 + r_2 = 5h - 2r_2$ $\Rightarrow 5h - 3r_2 = 5$$-1 = \frac{5k + r_2(3)}{5 + r_2}$ $\Rightarrow -5 - r_2 = 5k + 3r_2$ $\Rightarrow 5k + 4r_2 = -5$Since $C$ passes through $D(4, 0)$,the distance $BD = r_2$,so $(h-4)^2 + (k-0)^2 = r_2^2$.From the collinearity,the vector $\vec{AO}$ is parallel to $\vec{OB}$. $\vec{AO} = (1 - (-2), -1 - 3) = (3, -4)$.Since $O$ is $(1, -1)$ and $B$ is $(h, k)$,$\vec{OB} = (h-1, k+1)$.Since $\vec{OB} = \frac{r_2}{5} \vec{AO}$,we have $h-1 = \frac{3r_2}{5}$ and $k+1 = \frac{-4r_2}{5}$.$h = 1 + \frac{3r_2}{5}$ and $k = -1 - \frac{4r_2}{5}$.Substitute into $(h-4)^2 + k^2 = r_2^2$:$(\frac{3r_2}{5} - 3)^2 + (-1 - \frac{4r_2}{5})^2 = r_2^2$$\frac{9r_2^2}{25} - \frac{18r_2}{5} + 9 + 1 + \frac{8r_2}{5} + \frac{16r_2^2}{25} = r_2^2$$r_2^2 - 2r_2 + 10 = r_2^2$ $\Rightarrow 2r_2 = 10$ $\Rightarrow r_2 = 5$.

If a circle $C$ passing through $(4, 0)$ touches the circle $x^2 + y^2 + 4x - 6y - 12 = 0$ externally at a point $(1, -1),$ then the radius of the circle $C$ is

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