Inversion in a circle

Remarks

In coordinates with $O$ at the origin, the inversion is $x \mapsto r^2 x / \norm{x}^2$. In complex notation with center $0$, it is $z \mapsto r^2 / \bar{z}$. Adding a point $\infty$ and setting $O \mapsto \infty$ and $\infty \mapsto O$ extends it to a bijection of the Riemann sphere. It is an involution: applying it twice gives the identity. The same formula defines inversion in a sphere of $\R^n$.

Example: for the circle with center $(0, 0)$ and radius $1$, the point $P = (2, 0)$ goes to $P' = (1/2, 0)$. Check: $P'$ lies on the ray from the origin through $P$, and $\abs{OP} \cdot \abs{OP'} = 2 \cdot \frac{1}{2} = 1$.

Non-example: the map $x \mapsto -r^2 x / \norm{x}^2$ satisfies clause 2 but sends $P$ to the opposite ray, so clause 1 fails.