Let the tangents drawn from the origin to the circle, $x^{2}+y^{2}-8 x-4 y+16=0$ touch it at the points $A$ and $B$. The $(A B)^{2}$ is equal to :
$\frac{52}{5}$
$\frac{32}{5}$
$\frac{56}{5}$
$\frac{64}{5}$
Correct Option: , 4
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