Q:

The line $lx+my=1$ intersects the circle $x^2+y^2=a^2$ at points $A$, $B$ . . .

Problem: 

The line $lx+my=1$ intersects the circle $x^2+y^2=a^2$ at points $A$, $B$. If $AB$ subtends $\frac{\pi}{4}$ at origin then find $a^2(l^2+m^2)$ 

My approach: 
Can we find point of intersection of line and circle by putting $x = \frac{1-my}{l}$? But this way I am not getting the answer. Please suggest how to proceed in this. Thanks.

A:

Homogenize the line with the circle:
$$x^2+y^2=a^2(lx+my)^2$$
Rearrange to get the equation of pair of lines:
$$x^2(a^2l^2-1)+y^2(a^2m^2-1)+2a^2lmxy=0\tag{i}$$

The angle between the line pair $px^2+2qxy+ry^2=0$ is given by $\pm\tan\theta=\dfrac{2\sqrt{q^2-pr}}{p+r}$

In this case $\tan\theta=\pm1$, can you take it from here?

