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(A) The given circle is $C_1: x^2+y^2-4x-6y-12=0$. Its center $O_1$ is $(2, 3)$ and radius $R_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = 5$.Let the re

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$2$

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(A) The given circle is $C_1: x^2+y^2-4x-6y-12=0$. Its center $O_1$ is $(2, 3)$ and radius $R_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = 5$.Let the required circle be $C_2$ with center $O_2(h, k)$ and radius $R_2 = 3$.Since $C_2$ touches $C_1$ internally at $P(-1, -1)$,the center $O_2$ lies on the line joining $O_1(2, 3)$ and $P(-1, -1)$.The distance $O_1O_2 = R_1 - R_2 = 5 - 3 = 2$.Using the section formula,$O_2$ divides $O_1P$ externally in the ratio $R_1 : R_2 = 5 : 3$.$O_2 = \left( \frac{5(-1) - 3(2)}{5-3}, \frac{5(-1) - 3(3)}{5-3} \right) = \left( \frac{-11}{2}, -7 \right)$.The equation of $C_2$ is $(x + \frac{11}{2})^2 + (y + 7)^2 = 3^2$.$x^2 + 11x + \frac{121}{4} + y^2 + 14y + 49 = 9$.$x^2 + y^2 + 11x + 14y + \frac{121}{4} + 40 = 0$.$x^2 + y^2 + 11x + 14y + \frac{281}{4} = 0$.Here $p=11, q=14, r=\frac{281}{4}$.$p+q-r = 11+14-\frac{281}{4} = 25 - 70.25 = -45.25$. Wait,checking the point $(-1, -1)$ in the circle equation: $(-1)^2+(-1)^2+p(-1)+q(-1)+r = 0$ $\Rightarrow 2-p-q+r=0$ $\Rightarrow p+q-r=2$.

If the equation of the circle whose radius is $3$ units and which touches internally the circle $x^2+y^2-4x-6y-12=0$ at the point $(-1,-1)$ is $x^2+y^2+px+qy+r=0$,then $p+q-r=$

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