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

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

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

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

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

answer is D.

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Detailed Solution

$x^{2} + y^{2} - 4 x + 6 y - 37 = 0 \Rightarrow {(x - 2)}^{2} + {(y + 3)}^{2} = 50 x^{2} + y^{2} - 4 x + 6 y - 30 = 0 \Rightarrow {(x - 2)}^{2} + {(y + 3)}^{2} = 33 \begin{matrix} C_{1} = (2, - 3) \\ r_{1} = \sqrt{4 + 9 + 37} \\ r_{1} = \sqrt{50} \end{matrix} \begin{matrix} C_{2} = (2, - 3) \\ r_{2} = \sqrt{4 + 9 + 20} \\ r_{2} = \sqrt{33} \end{matrix} G i v e n c i r c l e s a r e c o n c e n t r i c c i r c l e s w i t h c e n t r e (2, - 3) Locus is {(x - 2)}^{2} + {(y + 3)}^{2} = {r_{1}}^{2} + {r_{2}}^{2} \Rightarrow x^{2} + y^{2} - 4 x + 6 y - 70 = 0$