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The common tangent to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} + 6 x + 8 y - 24 = 0$ also passes through the point

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$(6, - 2)$

$(4, - 2)$

$(- 6, 4)$

$(- 4, 6)$

answer is A.

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Detailed Solution

$S_{1} ≅ x^{2} + y^{2} - 4 = 0$ and

$S_{2} \equiv x^{2} + y^{2} + 6 x + 8 y - 24 = 0$

$∴ C_{1} (0, 0); r_{1} = 2$ and $C_{2} (- 3, - 4); r_{2} = 7$

$∵ C_{1} C_{2} = 5 = |r_{1} - r_{2}|$

$∵$ Circles touch each other internally.

$∵$ Common tangent will be $S_{1} - S_{2} = 0$

$= 6 x + 8 y - 20 = 0 \Rightarrow 3 x + 4 y = 10$

Only point in option (1), i.e., (6,-2) satisfies it.

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