Locus of the points of intersection of perpendicular tangents drawn, one to each of the circles $x^{2} + y^{2} - 4 x + 6 y - 37 = 0, x^{2} + y^{2} - 4 x + 6 y - 20 = 0$ is

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

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x^{2} + y^{2} - 4 x + 6 y - 50 = 0

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x^{2} + y^{2} - 4 x + 6 y - 57 = 0

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x^{2} + y^{2} - 4 x + 6 y - 70 = 0

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Solution

The correct option is D $x^{2} + y^{2} - 4 x + 6 y - 70 = 0$
Given two circles are concentric
$C_{1} : (x - 2)^{2} + (y + 3)^{2} = 50$
$C_{2} : (x - 2)^{2} + (y + 3)^{2} = 33$
Let $P (h, k)$ be the point of intersection of perpendicular tangents, one tangent drawn to each.
Refer to the figure.
$P A ⊥ C A, P B ⊥ P A, B C ⊥ P B$
Hence, $P A C B$ is a rectangle.
$P C^{2} = P A^{2} + A C^{2}$
$(h - 2)^{2} + (k + 3)^{2} = r_{1}^{2} + r_{2}^{2} = 83$
$\Rightarrow h^{2} + k^{2} - 4 h + 6 k - 70 = 0$
$\Rightarrow x^{2} + y^{2} - 4 x + 6 y - 70 = 0$
Hence, option D is the equation of locus.

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