Theorem. The stereographic projection preserves angles and circles.

Proof.

(i) Preserving angles.

Given a point $\mathbf{x_0}$ on the hemisphere $S^+$ and arbitrary vectors $a$, $b$ on its tangential plane $\mathcal{T}_0$ at $\mathbf{x_0}$ which enclose an angle $\varphi_0 = \varphi_0^a - \varphi_0^b$. Since every tangent to an arbitrary curve on $S^+$ at $\mathbf{x_0}$ is the tangent to a circle on $S^+$, $\varphi_0$ is the angle enclosed by the two circles, or their planes respectively. If we now normalize the coordinate system by means of a unitary transformation matrix $T$, such that in new coordinates $\mathbf{x_0}'$ coincides with $[ 0 \: 0 \: R ]^T$ and the normal of the circle's plane according to $a$ is parallel to $\mathbf{e}_{x'}$, we get for the projected angle in the equatorial plane $\varphi_0^E$: $$\cos \varphi_0^E = \frac{(T\mathbf{p}_a(r)) . (T\mathbf{p}_b(r))}{\lVert \mathbf{p}_a(r) \rVert \lVert \mathbf{p}_b(r) \rVert} = \frac{\mathbf{p}_a(r)^T T^T T \mathbf{p}_b(r)}{\lVert \mathbf{p}_a(r) \rVert \lVert \mathbf{p}_b(r) \rVert} = \frac{\mathbf{p}_a(r)^T \mathbf{p}_b(r)}{\lVert \mathbf{p}_a(r) \rVert \lVert \mathbf{p}_b(r) \rVert} = \cos(\varphi_0^a)\cos(\varphi_0^b) + \sin(\varphi_0^a)\sin(\varphi_0^b) = \cos(\varphi_0^a - \varphi_0^b),$$ where $$\mathbf{p}_i(r) = \left( \begin{array}{c} \cos \varphi_i \\ \sin \varphi_i \\ 0 \end{array} \right) \frac{\partial}{\partial r} \left( \frac{r}{1 + \sqrt{1 - \left( \frac{r}{R} \right)^2 }} \right) \qquad \text{and} \qquad \frac{\mathbf{p}_i(r)}{\lVert \mathbf{p}_i(r) \rVert} = \left( \begin{array}{c} \cos \varphi_i \\ \sin \varphi_i \\ 0 \end{array} \right), \qquad i \in \{a,b\}$$ is the derivative of the stereographic projection of a $r$-parameterized curve on $S^+$ and its direction, respectively.

(ii) Preserving circles.

Given an arbitrary circle on $S^+$, parameterized over $\varphi$, an arbitrary but constant radius $r_0$ and a unitary transformation $T$ such that the normal of the circle's plane is parallel to the equatorial plane, we have, for the projection of the circle $$\mathbf{p}(r) = T \left( \begin{array}{c} \cos \varphi \\ \sin \varphi \\ 0 \end{array} \right) \frac{r_0}{1 + \sqrt{1 - \left( \frac{r_0}{R} \right)^2 }}$$ Since a unitary transformation perserves the circle's properties (see also (i)), $\mathbf{p}(r)$ describes a scaled circle in the equatorial plane.
2015 JAN 02 (v1.0)
Contact me: m.herrmann followed by an -at- followed by blaetterundsterne.org.

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