The number of common tangents to the circles | vedclass

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(B) Given circles are $S_1 \equiv x^2 + y^2 = 2^2$ and $S_2 \equiv x^2 + y^2 - 6x - 8y - 24 = 0$.For $S_2$,the centre is $(3, 4)$ and radius $r_2 = \s

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$1$

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(B) Given circles are $S_1 \equiv x^2 + y^2 = 2^2$ and $S_2 \equiv x^2 + y^2 - 6x - 8y - 24 = 0$.For $S_2$,the centre is $(3, 4)$ and radius $r_2 = \sqrt{3^2 + 4^2 - (-24)} = \sqrt{9 + 16 + 24} = \sqrt{49} = 7$.For $S_1$,the centre is $C_1 = (0, 0)$ and radius $r_1 = 2$.The distance between the centres $C_1(0, 0)$ and $C_2(3, 4)$ is $d = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16} = 5$.Since $d = |r_2 - r_1| = |7 - 2| = 5$,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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