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Let the tangents drawn from the origin to the circle $x^{2} + y^{2} - 8 x - 4 y + 16 = 0$ touch it at the point A and B, then ${(A B)}^{2}$ is equal to

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$\frac{56}{5}$

$\frac{32}{5}$

$\frac{64}{5}$

$\frac{52}{5}$

answer is D.

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Detailed Solution

$C = (4, 2)$

$R = \sqrt{16 + 4 - 16} = 2$

$L = \sqrt{S_{11}} = \sqrt{16} = 4$

$A B = \frac{2 L R}{\sqrt{L^{2} + R^{2}}} = \frac{16}{\sqrt{20}}$

$A B^{2} = \frac{64}{5}$

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