If the chord y=mx+1 of the circle x^2+y^2=1 subtends an angle of measure 45^ at the major …

Question

If the chord $y=mx+1$ of the circle $x^2+y^2=1$ subtends an angle of measure $45^\circ$ at the major segment of the circle, then the value of $m$ is:

Accepted Answer

Homogenizing $x^2+y^2=1$ with $y-mx=1$ (i.e. $1=(y-mx)^2$):
$$x^2+y^2=(y-mx)^2\Rightarrow x^2(1-m^2)+2mxy=0.$$
This pair of lines through the origin has angle $45^\circ$ between them:
$$\tan45^\circ=1=\left|\frac{2\sqrt{m^2-0}}{1-m^2}\right|=\frac{2|m|}{|1-m^2|}\Rightarrow 1-m^2=\pm2m\Rightarrow m^2\pm2m-1=0.$$
Solving: $m=\dfrac{-2\pm\sqrt{4+4}}{2}=-1\pm\sqrt2.$

Solutions · 1