Locus of point of intersection of perpendicular tangents to the circle $x^{2} + y^{2} - 4 x - 6 y - 1 = 0$ is

x^{2} + y^{2} - 4 x - 6 y - 15 = 0

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x^{2} + y^{2} - 4 x - 6 y + 15 = 0

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x^{2} + y^{2} - 4 x - 3 y - 15 = 0

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x^{2} + y^{2} + 4 x + 6 y - 15 = 0

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Solution

The correct option is C $x^{2} + y^{2} - 4 x - 6 y - 15 = 0$
Locus of intersection of perpendicular tangents is the director circle.
Here circle is $\to x^{2} + y^{2} - 4 x - 6 y - 1 = 0$
$\Rightarrow (x - 2)^{2} + (y - 3)^{2} - 14 = 0$
$\Rightarrow (x - 2)^{2} + (y - 3)^{2} = 14$
$∴$ Director circle:
$(x - 2)^{2} + (y - 3)^{2} = 2 (14)$
$\Rightarrow (x - 2)^{2} + (y - 3)^{2} = 28$
$\Rightarrow x^{2} + y^{2} - 4 x - 6 y + 13 = 28$
$\Rightarrow x^{2} + y^{2} - 4 x - 6 y - 15 = 0$
$∴ A)$ Answer.

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