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A line $l$ meets the circle $x^{2} + y^{2} = 61$ in A,B and P(-5,6) is such that PA=PB=10 then the equation of ‘ $l$ ’ is

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$5 x + 6 y + 11 = 0$

$5 x - 6 y - 11 = 0$

$5 x - 6 y + 11 = 0$

$5 x - 6 y + 12 = 0$

answer is C.

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Detailed Solution

$\begin{array}{l} S = {(x + 5)}^{2} + {(y - 6)}^{2} = 100 \\ S^{1} \equiv x^{1} + y^{2} = 61 \\ l = s - s^{1} = 0 \end{array}$

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