Circles Riemann sphere

Question

I was able to prove that the circle passing through the north pole of the Riemann sphere is mapped into a straight line. How can I prove that the circle on the Riemann sphere corresponds to a circle on the complex plane. Thank you in advance!

Answer 1

A complex number $x+iy$ corresponds to the point $(u,v,w)$ on the Riemann sphere with $$ u=\frac{2x}{r^2+1},\quad v=\frac{2y}{r^2+1},\quad w=\frac{r^2-1}{r^2+1},\qquad\text{where $r=\sqrt{x^2+y^2}$}. $$ A circle on the Riemann sphere is the intersection of the sphere with a plane. Write the equation of the plane as $au+bv+cw=d$. Substitute in the above coordinates and multiply by the common denominator $r^2+1$, substitute in $r^2=x^2+y^2$, and you get a quadratic equation for a circle (unless $c=d$, when the plane passes through the north pole of the Riemann sphere (a.k.a. infinity)). I leave the details up to you.