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

$4 x^{2} + 4 y^{2} - 16 x + 13 y = 0$

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$x^{2} + y^{2} - 17 x + 15 y = 0$

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$4 x^{2} + 4 y^{2} - 16 x + 12 y = 0$

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$3 x^{2} + 3 y^{2} - 17 x - 15 y = 0$

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Solution

The correct option is C
$4 x^{2} + 4 y^{2} - 16 x + 12 y = 0$

Given circle is,

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

and external point is (2, 2) we have to use the formula

$S S_{1} = T^{2}$

$[x^{2} + y^{2} - 4 x + 4 y] [4 + 4 - 8 + 8] = 2 x + 2 y - 2 (x + 2) + 2 (y + 2)$

$8. [x^{2} + y^{2} - 4 x + 4 y] = 2 y + 2 y - 4 + 4$

$4. [x^{2} + y^{2} - 4 x + 4 y] = 4 y$

$4 x^{2} + 4 y^{2} - 16 x + 12 y = 0$

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