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

Normal at a point to the circle intersect the circle again at the other end of the diameter passing along the normal,

Centre of the circle $= C (3, - 4) .$ Given point $A = (8, 8)$

Let B be the point at with the normal meets the circle again.

Then midpoint of $A B = C \Rightarrow B = 2 C - A = 2 (3, - 4) - (8, 8) = (- 2, - 16)$ .

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