Question:
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:
Correct Option: , 3
Solution:
$L=\sqrt{S_{1}}=\sqrt{16}=4$
$R=\sqrt{16+4-16}=2$
Length of chord of contact
$=\frac{2 L R}{\sqrt{L^{2}+R^{2}}}=\frac{2 \times 4 \times 2}{\sqrt{16+4}}=\frac{16}{\sqrt{20}}$
quare of length of chord of contact $=\frac{64}{5}$
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