# Radius of the inverse image of a circle under stereographic projection

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.

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## 2 Answers

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 $$\rho=\sin\phi={\tan\phi\over\sqrt{1+\tan^2\phi}}={2r\over\sqrt{(1+a^2)^2+2r^2+r^4}}\ .$$

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Let $\Pi$ be the plane whose intersection with the unit sphere $S$ is the spherical image of the circle $C$ in the plane with center $a$ and radius $r$, i.e., $\Pi \cap S = \sigma^{-1}(C)$. We can write the equation of $\Pi$ implicitly in the form ($\alpha_i \in \mathbb{R}$) (Ahlfors, p.19)

$$\alpha_1 x_1+\alpha_2 x_2 + \alpha_3 x_3 = \alpha_0, \quad \alpha_1 + \alpha_2 + \alpha_3 = 1 \quad 0 \leq \alpha_0 < 1,$$

where $(x_1, x_2, x_3)$ are coordinate functions of the ambient space that the sphere $S$ lives in.

The circle $C$ corresponding to the intersection of this plane $\Pi$ and the unit sphere $S$ is given by (Ahlfors, p.19)

$$x^2 + y^2 - 2\frac{\alpha_1}{\alpha_0-\alpha_3}x+2\frac{\alpha_2}{\alpha_0-\alpha_3}y + \frac{\alpha_0}{\alpha_0-\alpha_3} - \frac{\alpha_3}{\alpha_0-\alpha_3} = 0$$

where $x$, $y$ are the real and imaginary parts of $z$, respectively. On the other hand, our circle has the given equation $\lvert z - a\rvert= r$, which written in coordinates $z = x + iy$ and $a = (a_1, a_2)$ has the form

$$(x-a_1)^2 + (y-a_2)^2 = r^2 \Rightarrow x^2 + y^2 - 2 a_1 x - 2 a_2 y + a_1^2 + a_2^2 - r^2 = 0$$

Equating the like powers of these two last equations, we can solve for $\alpha_0$, in particular, which turns out to be

$$\alpha_0 = \pm \frac{1 + \lvert a \rvert^2 - r^2}{\sqrt{\lvert a \rvert^4 - 2\lvert a \rvert^2(1-r^2)+(1+r^2)^2}}$$

Now, our sphere $S$ is the unit sphere (the set of points that is distance 1 from the origin of $\mathbb{R}^3$) and the number $\alpha_0$ we just computed is the length of the line from the origin of $\mathbb{R}^3$ to the center of the circle $\sigma^{-1}(C)$ as provided in the Wikipedia article:

http://en.wikipedia.org/wiki/Plane%E2%80%93sphere_intersection

Thus from Pythagorean theorem, the desired radius $\rho$ of the circle $\sigma^{-1}(C)$ is given by

$$\boxed{ \rho = \sqrt{1-\alpha_0^2} = \Large{ \frac{2r}{\sqrt{\lvert a \rvert^4 - 2\lvert a \rvert^2(1-r^2)+(1+r^2)^2}}} }$$

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