The lengths of the tangents from the points $A$and $B$ to a circle are $l_1$and $l_2$respectively. If points are conjugate with respect to the circle, then $A{B}^2=$

Let the circle be $x^2+y^2=a^2$ and let $A\left(x_1, y_1\right)$ $B\left(x_2, y_2\right)$ be the given points. Then,

$l_1^2=x_1^2+y_1^2-a^2,l_2^2=x_2^2+y_2^2-a^2$

If the  points $A\left(x_1,y_1\right)$  and B ($x_{2,}{y}_2)$    with respect to $x^2+y^2=a^2$  then 

 $x_1{x}_2+y_1{y}_2=a^2$$\to \left(i\right)$

now 

$\begin{array}{l}A{B}^2={\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2\\ \Rightarrow A{B}^2=\left(x_1^2+y_1^2\right)+\left(x_2^2+y_2^2\right)-2\left(x_1{x}_2+y_1{y}_2\right)\end{array}$

$\Rightarrow A{B}^2=\left({x_1}^2+{y_1}^2\right)+\left({x_2}^2+{y_2}^2\right)-2a^2\left[\text{ Using (i) }\right]$

$\Rightarrow A{B}^2=\left(x_1^2+y_1^2-a^2\right)+\left(x_2^2+y_2^2-a^2\right)=l_1^2+l_2^2$