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Tangents from point ‘P’ are drawn one to each of circle $x^{2} + y^{2} - 2 x + 6 y + 1 = 0$ ; $x^{2} + y^{2} - 2 x + 6 y - 3 = 0$ if the tangents are perpendicular then locus of P is

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$x^{2} + y^{2} + 2 x - 6 y - 12 = 0$

$x^{2} + y^{2} - 2 x - 6 y - 12 = 0$

$x^{2} + y^{2} - 2 x + 6 y + 12 = 0$

$x^{2} + y^{2} - 2 x + 6 y - 12 = 0$

answer is D.

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Detailed Solution

centre = $(1, - 3)$

$r_{1} = \sqrt{1 + 9 - 1} = 3$

$r_{2} = \sqrt{1 + 9 + 3} = \sqrt{13}$

${(x - 1)}^{2} + {(y + 3)}^{2} = r_{1}^{2} + r_{2}^{2}$

$\Rightarrow x^{2} + y^{2} - 2 x + 6 y + 1 + 9 = 9 + 13$

$\Rightarrow$ $x^{2} + y^{2} - 2 x + 6 y - 12 = 0$

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