The number of common tangents to the circles | vedclass

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(C) For the circle $x^2+y^2-4x-6y-12=0$,the center $C_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$. For the cir

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$3$

Vedclass · Answer

(C) For the circle $x^2+y^2-4x-6y-12=0$,the center $C_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$. For the circle $x^2+y^2+6x+18y+26=0$,the center $C_2 = (-3, -9)$ and radius $r_2 = \sqrt{(-3)^2+(-9)^2-26} = \sqrt{9+81-26} = \sqrt{64} = 8$. The distance between the centers $d = C_1C_2 = \sqrt{(2-(-3))^2 + (3-(-9))^2} = \sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13$. Since $r_1 + r_2 = 5 + 8 = 13$,we have $d = r_1 + r_2$. Because the distance between the centers is equal to the sum of the radii,the two circles touch each other externally. When two circles touch each other externally,there are exactly $3$ common tangents.

The number of common tangents to the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+18y+26=0$ is:

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