Given two circles x^2+y^2+8x-6y-24=0 and x^2+ | vedclass

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(D) For the first circle $x^2+y^2+8x-6y-24=0$,the center $C_1 = (-4, 3)$ and radius $r_1 = \sqrt{(-4)^2 + 3^2 - (-24)} = \sqrt{16+9+24} = \sqrt{49} =

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Touching externally.

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(D) For the first circle $x^2+y^2+8x-6y-24=0$,the center $C_1 = (-4, 3)$ and radius $r_1 = \sqrt{(-4)^2 + 3^2 - (-24)} = \sqrt{16+9+24} = \sqrt{49} = 7$. For the second circle $x^2+y^2-4x+10y+20=0$,the center $C_2 = (2, -5)$ and radius $r_2 = \sqrt{2^2 + (-5)^2 - 20} = \sqrt{4+25-20} = \sqrt{9} = 3$. The distance between the centers $C_1 C_2 = \sqrt{(2 - (-4))^2 + (-5 - 3)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36+64} = \sqrt{100} = 10$. Since $C_1 C_2 = r_1 + r_2 = 7 + 3 = 10$,the two circles touch each other externally.

Given two circles $x^2+y^2+8x-6y-24=0$ and $x^2+y^2-4x+10y+20=0$. Then they are

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