The number of common tangents of the circles | vedclass

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(A) For the first circle $x^2+y^2=4$,the center $C_1 = (0,0)$ and radius $r_1 = \sqrt{4} = 2$. For the second circle $x^2+y^2-6x-8y-24=0$,the center $

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$1$

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(A) For the first circle $x^2+y^2=4$,the center $C_1 = (0,0)$ and radius $r_1 = \sqrt{4} = 2$. For the second circle $x^2+y^2-6x-8y-24=0$,the center $C_2 = (3,4)$ and radius $r_2 = \sqrt{3^2+4^2-(-24)} = \sqrt{9+16+24} = \sqrt{49} = 7$. The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(3-0)^2+(4-0)^2} = \sqrt{9+16} = 5$. We observe that $|r_2 - r_1| = |7 - 2| = 5$. Since the distance between the centers $d = |r_2 - r_1|$,the circles touch each other internally. When two circles touch each other internally,they have exactly $1$ common tangent.

The number of common tangents of the circles $x^2+y^2-4=0$ and $x^2+y^2-6x-8y-24=0$ is

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