Stereographic projection preserves circles

Statement

Setup

Further information: Stereographic projection

Consider ${\mathbb{R}}^3$, three-dimensional Euclidean space, with coordinates $x,y,z$. Denote:

$S^2:=\{(x,y,z)\in {\mathbb{R}}^3\mid x^2+y^2+z^2=1\}$

Identify $\mathbb{C}$ with the $xy$-plane under the map:

$(x,y,0)\mapsto x+iy$

Let $N=(0,0,1)$ denote the north pole in $S^2$, and define the stereographic projection as a bijective map:

$S^2\setminus \{N\}\to \mathbb{C}$

which sends a point $P\in S^2\setminus \{N\}$ to the unique point in $\mathbb{C}$ that is collinear with $N$ and $P$.

Actual statement

The following two facts are true:

• Circles in $S^2\setminus \{N\}$ get mapped to circles in $\mathbb{C}$
• A circle in $S^2$ that passes through $N$, minus the point $N$, gets mapped to a straight line in $\mathbb{C}$

Facts used

• Distance formula for inverse stereographic projection

Proof

Computational proof

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