I need to find the radius of the circle on the Riemann sphere $S$ whose stereographic projection is $C(a;r)$, i.e. the circle with centre $a$ and radius $r$ in the complex plane.
I have observed that, if two diametrically opposite points on the circle $C(a;r)$ have also their preimage as two diametrically opposite points of the circle on $S$, then one can compute the required radius which is $\frac{1}{2}\sigma (|a|+r,|a|-r)$ (where $\sigma$ is the metric corresponding to stereographic projection). It is kind of obvious, but I am unable to prove this fact.
Best Answer
The stereographic projection $\sigma$ commutes with rotations around the vertical axis. It follows that the euclidean radius $\rho$ of the circle $\gamma:=\sigma^{-1}(C)\subset S$ depends only on $|a|$ and $r$. So let's assume $a>0$.
Draw a figure where the plane $\Pi$ containing the circle $\gamma$ appears as a line. This line intersects the unit circle in two points $P_1$, $\,P_2$, and the stereographic images of these two points are the points $a+r$ and $a-r$ on the horizontal axis. The ray from the north pole $N$ to $P_1$, resp. $a+r$ includes an angle $\alpha>0$ with the vertical axis, and the ray from $N$ to $P_2$, resp. $a-r$ includes an angle $\beta$ with the vertical axis, where $\beta<0$ if $a-r<0$. Let $\phi:=\alpha-\beta\geq0$ be the angle between these two rays. Then $$\tan\phi={\tan\alpha -\tan\beta\over 1+\tan\alpha\,\tan\beta}={(a+r)-(a-r)\over 1+(a+r)(a-r)}={2r\over 1+a^2-r^2}\ .$$ The two rays intersect the unit circle in the two points $P_1$ and $P_2$, and one has $$\angle(P_1OP_2)=2\,\angle(P_1NP_2)=2\phi\ .$$ It follows that the radius $\rho$ of the circle $\gamma$ is given by $$\rho=\sin\phi={\tan\phi\over\sqrt{1+\tan^2\phi}}={2r\over\sqrt{(1+a^2)^2+2r^2+r^4}}\ .$$