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If the length of the tangent drawn from any point on the circle $x^{2} + y^{2} + 15 x - 17 y + c^{2} = 0$ to the circle $x^{2} + y^{2} + 15 x - 17 y + 21 = 0$ is $\sqrt{5}$ units, Then $c^{2} - 12$ is equal to

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$- 4$

answer is D.

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Detailed Solution

$\sqrt{C - C^{'}} = \sqrt{5} \dots \dots \dots \sqrt{21 - C^{2}} = \sqrt{5}, 21 - c^{2} = 5 \Rightarrow c^{2} = 21 - 5 C^{2} - 12 = 4$

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