Let \(\displaystyle A(x_1,y_1), \ B(x_1,y_1)\) be the intersect points between the line and the circle. Then \(\displaystyle \widehat{AOB}=45^{\circ}\).
The line AO has the slope \(\displaystyle m_1=\frac{y_1}{x_1}\) and the line BO has the slope \(\displaystyle m_2=\frac{y_2}{x_2}\)
The tangent of the acute angle between the two lines is \(\displaystyle \tan\widehat{AOB}=\left|\displaystyle\frac{m_1-m_2}{1+m_1m_2}\right|\)
Replace \(\displaystyle m_1\) and \(\displaystyle m_2\):
\(\displaystyle 1=\tan 45^{\circ}=\left|\displaystyle\frac{x_2y_1-x_1y_2}{x_1x_2+y_1y_2}\right|\), (1)
The coordinates of A and B are the solutions of the system formed by the equation of the line and equation of the circle.
\(\displaystyle y=\displaystyle\frac{1-lx}{m}\).
Replace y in the equation of the circle and we have:
\(\displaystyle (m^2+l^2)x^2-2lx+1-a^2m^2=0\)
\(\displaystyle x_1, \ x_2\) are the roots of the quadratic and we have
\(\displaystyle x_1+x_2=\displaystyle\frac{2l}{m^2+l^2}, \ x_1x_2=\displaystyle\frac{1-a^2m^2}{m^2+l^2}\)
Replace \(\displaystyle y_1,y_2\) in (1):
\(\displaystyle 1=\displaystyle\frac{|m|\cdot|x_2-x_1|}{|(m^2+l^2)x_1x_2-l(x_1+x_2)+1|}=\frac{|m|\sqrt{(x_1+x_2)^2-4x_1x_2}}{|(m^2+l^2)x_1x_2-l(x_1+x_2)+1|}\).
Now replace \(\displaystyle x_1+x_2\) and \(\displaystyle x_1x_2\).
\(\displaystyle 1=\displaystyle\frac{2\sqrt{a^2m^2+a^2l^2-1}}{2-a^2m^2-a^2l^2}\Rightarrow 2\sqrt{a^2m^2+a^2l^2-1}=1-(a^2m^2+a^2l^2-1)\)
Let \(\displaystyle \sqrt{a^2m^2+a^2l^2-1}=t\).
Then \(\displaystyle 2t=1-t^2\Rightarrow t^2+2t-1=0\)
The positive root is \(\displaystyle t=\sqrt{2}-1\).
\(\displaystyle \sqrt{a^2m^2+a^2l^2-1}=\sqrt{2}-1\).
Square both members:
\(\displaystyle a^2m^2+a^2l^2-1=3-2\sqrt{2}\Rightarrow a^2(m^2+l^2)=4-2\sqrt{2}\)