Purely Geometric Proof that Circles Invert to Circles

Question

I recently watched Douglas Hofstadter recall at the beginning of a talk about Feuerbach's Theorem the fact that under circle inversion, circles (which do not pass through the center of inversion) map to circles. He claimed it used the fact that if two inscribed angles cut the same arc, they must have equal measure (chord-angle theorem).

I tried to rediscover the proof, but concluded that his chord-angle theorem was useless. Assuming his definition of circle inversion indeed inverts distances from the center, he must also use some result about distance, but power of a point proves circle inversion easily without direct reference to the chord-angle theorem.

Is there a well known proof involving the chord-angle theorem?

Answer 1

Here's a straightforward proof of the circles-invert-to-circles theorem, where the linchpin is the Inscribed Angle Theorem (more-specifically, Thales' Theorem: An angle inscribed in a semicircle is a right angle.).

Consider a point $C$ on some circle. We wish to invert $C$ in a circle with center $O$. Let $\overline{AB}$ be the diameter of $C$'s circle such that $O$, $A$, $B$ are collinear. Let the inversions of $A$, $B$, $C$ be $A^\prime$, $B^\prime$, $C^\prime$.

Recall that the nature of inversion is that $$|\overline{OX}||\overline{OX^\prime}| = \left(\;\text{radius of $\bigcirc O$}\;\right)^2 \tag{1}$$ The radius is irrelevant; what matters is the constant product, so that we have $$|\overline{OA}||\overline{OA^\prime}| = |\overline{OB}||\overline{OB^\prime}| = |\overline{OC}||\overline{OC^\prime}| \tag{2}$$

This allows us to argue as follows: $$\begin{align} \frac{|\overline{OA}|}{|\overline{OC}|} = \frac{|\overline{OC^\prime}|}{|\overline{OA^\prime}|} &\implies \triangle OAC \sim \triangle OC^\prime A^\prime \implies \angle A^\prime C^\prime C = \angle A \tag{3a} \\[8pt] \frac{|\overline{OB}|}{|\overline{OC}|} = \frac{|\overline{OC^\prime}|}{|\overline{OB^\prime}|} &\implies \triangle OBC \sim \triangle OC^\prime B^\prime \implies \angle B^\prime C^\prime O = \angle B \tag{3b} \end{align}$$

Consequently, $$\angle ACB \cong \angle A^\prime C^\prime B^\prime \tag{4}$$

Now, Thales both sets us up and brings us home ...

$$\begin{align} \text{$C$ lies on a semicircle on $\overline{AB}$}\quad &\implies\quad \text{$\angle ACB$ is a right angle} \\[6pt] &\implies\quad \text{$\angle A^\prime C^\prime B^\prime$ is a right angle} \\ &\implies\quad \text{$C^\prime$ lies on a semicircle on $\overline{A^\prime B^\prime}$} \end{align}$$

Thus, the points on a circle invert to points on another circle, and (with some handwaving about continuity and bijection) we're done. $\square$

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In point of fact, the restriction to Thales' Theorem and right angles and diameters is unnecessary. We can take $\overline{AB}$ to be any (non-degenerate) chord of $C$'s circle, so long as it's collinear with $O$, and the proof effectively works as-is, requiring only that we replace "right angle" and "semi-circle" with appropriately-generic descriptors.

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By request, here are diagrams for varying configurations. In all cases, $(2)$ leads to similar triangles that imply $(3a)$, $(3b)$, and, ultimately $(3c)$, so that $\angle A^\prime C^\prime B^\prime$ is a right angle. "Usually", this means that $C^\prime$ lies on a regular circle with diameter $\overline{A^\prime B^\prime}$; when $A=O$, we have that $A^\prime$ is the "point at infinity", and we see that $C^\prime$ lies on the perpendicular line (that is, a circle of infinite radius) through $B^\prime$.