The equation of pair of tangents drawn to the circle $x^{2} + y^{2} - 2 x + 4 y + 3 = 0$ from point $(6, - 5)$ is

7 x^{2} + 23 y^{2} + 30 x y + 66 x + 50 y - 73 = 0

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7 x^{2} + 23 y^{2} - 30 x y + 66 x - 50 y - 73 = 0

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7 x^{2} + 3 y^{2} + 30 x y - 66 x + 50 y - 73 = 0

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3 x^{2} + 7 y^{2} + 30 x y + 66 x + 50 y - 73 = 0

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Solution

The correct option is A $7 x^{2} + 23 y^{2} + 30 x y + 66 x + 50 y - 73 = 0$
$S : x^{2} + y^{2} - 2 x + 4 y + 3$
$S_{1} : (6)^{2} + (- 5)^{2} - 2 \times 6 + 4 \times (- 5) + 3 = 32$
$T : x (6) - y (5) + (- 1) (x + 6) + 2 (y - 5) + 3$
$= 6 x - 5 y - x - 6 + 2 y - 10 + 3$
$= 5 x - 3 y - 13$

Equation of pair of tangents is given by
$S S_{1} = T^{2}$
$\Rightarrow 32 (x^{2} + y^{2} - 2 x + 4 y + 3) = (5 x - 3 y - 13)^{2}$
$\Rightarrow 32 x^{2} + 32 y^{2} - 64 x + 128 y + 96 = 25 x^{2} + 9 y^{2} + 169 - 30 x y + 78 y - 130 x$
$\Rightarrow 7 x^{2} + 23 y^{2} + 30 x y + 66 x + 50 y - 73 = 0$

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