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(B) For the circle $x^2+y^2-4x+10y+20=0$,the center is $C_1 = (2, -5)$ and the radius is $r_1 = \sqrt{2^2 + (-5)^2 - 20} = \sqrt{4+25-20} = 3$. For th

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$3x-4y-11=0$

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(B) For the circle $x^2+y^2-4x+10y+20=0$,the center is $C_1 = (2, -5)$ and the radius is $r_1 = \sqrt{2^2 + (-5)^2 - 20} = \sqrt{4+25-20} = 3$. For the circle $x^2+y^2+8x-6y-24=0$,the center is $C_2 = (-4, 3)$ and the radius is $r_2 = \sqrt{(-4)^2 + 3^2 - (-24)} = \sqrt{16+9+24} = 7$. The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(2 - (-4))^2 + (-5 - 3)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36+64} = 10$. Since $d = r_1 + r_2$ $(10 = 3 + 7)$,the circles touch each other externally at a single point. The equation of the common tangent at the point of contact is the radical axis,given by $S_1 - S_2 = 0$. $(x^2+y^2-4x+10y+20) - (x^2+y^2+8x-6y-24) = 0$. $-12x + 16y + 44 = 0$. Dividing by $-4$,we get $3x - 4y - 11 = 0$.

The equation of the common tangent to the circles $x^2+y^2-4x+10y+20=0$ and $x^2+y^2+8x-6y-24=0$ is

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