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(D) The equations of the circles are $S_1: x^2+y^2-4x+6y+4=0$ and $S_2: x^2+y^2+2x-2y-2=0$. For $S_1$,the center $C_1 = (2, -3)$ and radius $r_1 = \sq

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$-1$

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(D) The equations of the circles are $S_1: x^2+y^2-4x+6y+4=0$ and $S_2: x^2+y^2+2x-2y-2=0$. For $S_1$,the center $C_1 = (2, -3)$ and radius $r_1 = \sqrt{2^2+(-3)^2-4} = \sqrt{4+9-4} = 3$. For $S_2$,the center $C_2 = (-1, 1)$ and radius $r_2 = \sqrt{(-1)^2+1^2-(-2)} = \sqrt{1+1+2} = 2$. The distance between the centers is $C_1C_2 = \sqrt{(2 - (-1))^2 + (-3 - 1)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = 5$. Since $r_1 + r_2 = 3 + 2 = 5 = C_1C_2$,the circles touch each other externally at point $P$. The point of contact $P$ divides the line segment $C_1C_2$ internally in the ratio $r_1:r_2 = 3:2$. $P = \left( \frac{3(-1) + 2(2)}{3+2}, \frac{3(1) + 2(-3)}{3+2} \right) = \left( \frac{-3+4}{5}, \frac{3-6}{5} \right) = \left( \frac{1}{5}, -\frac{3}{5} \right)$. The common tangent at $P$ is the radical axis of the two circles,given by $S_1 - S_2 = 0$. $(x^2+y^2-4x+6y+4) - (x^2+y^2+2x-2y-2) = 0$ $-6x + 8y + 6 = 0$ Dividing by $-2$,we get $3x - 4y - 3 = 0$. Comparing this with $ax+by+c=0$,we have $a=3, b=-4, c=-3$. Thus,$\frac{a}{c} = \frac{3}{-3} = -1$.

If the equation of the common tangent of the circles $x^2+y^2-4x+6y+4=0$ and $x^2+y^2+2x-2y-2=0$ at their point of contact is $ax+by+c=0$,then $\frac{a}{c}=$

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