# Let the tangents drawn from the origin to the circle,

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:

1. (1) $\frac{52}{5}$

2. (2) $\frac{56}{5}$

3. (3) $\frac{64}{5}$

4. (4) $\frac{32}{5}$

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 $=\frac{64}{5}$