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The normal to the circle given by $x^{2} + y^{2} - 6 x + 8 y - 144 = 0$ at $(8, 8)$ meets the circle again at the point

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$(2, - 16)$

$(2, 16)$

$(- 2, 16)$

$(- 2, - 16)$

answer is D.

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Detailed Solution

$y_{1} (x + g) - x_{1} (y + f) = 0 (x_{1}, y_{1}) = (8, 8) (g, f) = (- 3, 4) 8 (x - 3) - 8 (y + 4) = 0 x - 3 - y - 4 = 0 x - y - 7 = 0$

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