The number of common tangents to the circles | vedclass

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(B) For the circle $x^2 + y^2 = 4$,the center $C_1 = (0, 0)$ and radius $r_1 = 2$.For the circle $x^2 + y^2 - 6x - 8y - 24 = 0$,the center $C_2 = (3,

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(B) For the circle $x^2 + y^2 = 4$,the center $C_1 = (0, 0)$ and radius $r_1 = 2$.For the 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 compare $d$ with the sum and difference of the radii:$r_1 + r_2 = 2 + 7 = 9$$|r_1 - r_2| = |2 - 7| = 5$.Since $d = |r_1 - r_2|$,the two circles touch each other internally.When two circles touch internally,there is exactly $1$ common tangent.

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

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