Geometric transformation

Theorem: $T(C)$ and $T(g)$ are both circles.

It is easy to exclude that neither $C$ nor $g$ is transformed into a straight line. For each curve $\gamma$ we can define a distance to the origin by $d_\gamma = \text{min}_{\vec P \in \gamma}{| \vec P | }$ . Since the coordinate transformation inverts the lengths, the point closest to the origin is mapped to a point furthest from the origin. Thus, the image $T(\gamma)$ of curve $\gamma$ must be within a circle of radius $d_\gamma^{-1}$ , because $|\vec P'| = |\vec P|^{-1} \leq d_\gamma^{-1}$ for each $\vec P \in \gamma$ . Since neither $C$ nor $g$ goes through the origin, their distances are strictly positive ( $d_C, d_g > 0$ ), so that their images $T(C)$ and $T(g)$ are bounded. But a straight line continues to infinity, so that there is a contradiction here.

The proof that the curves are actually circles is much more complicated. In the proof here, which I have worked for myself, I use complex numbers to algebraically represent the transformation. But there may be other proofs that are much shorter or more elegant.

Proof:

We treat the vectors $\vec P = (x,y) \in \mathbb{R}^2$ as complex numbers $z = x + iy = r e^{i \varphi} \in \mathbb{C}$ with real part $x = r \cos \varphi$ and imaginary part $y = r \sin \varphi$ . Thus, the transformation $T$ corresponds to a mapping $f: z \mapsto \frac{1}{z^\ast} = \frac{1}{r} e^{i \varphi}$ Therefore, $f(z)$ and $z$ have the same phase $\varphi$ and point in the same direction. Furthermore, $|f(z)| = r^{-1} = |z|^{-1}$ , so that the absolute values are inverse to each other.

A circle around point $a$ with radius $R$ corresponds to the points $z = a + \frac{a}{|a|} R e^{i \varphi} = a \left(1 + \rho e^{i \varphi} \right), \quad \rho = \frac{R}{|a|}, \varphi \in [0,2\pi]$ We assume $\rho \not= 1$ (the origin is not intersected), so that $\begin{aligned} f(z) &= \frac{1}{a^\ast} \frac{1}{1 + \rho e^{-i \varphi}} \\ &= \frac{1}{a^\ast} \left( \frac{1}{1 - \rho^2} - \frac{1}{1 - \rho^2} + \frac{1}{1 + \rho e^{-i \varphi}} \right)\\ &= \frac{1}{a^\ast} \left( \frac{1}{1 - \rho^2} - \frac{\rho (\rho + e^{-i \varphi})}{(1 - \rho^2)(1 + \rho e^{-i \varphi})} \right)\\ &= \frac{1}{(1 - \rho^2) a^\ast} \left( 1 - \rho \frac{\rho + e^{- i \varphi}}{1 + \rho e^{- i \varphi}} \right)\\ &= a' + \frac{a'}{|a'|} R' e^{i \varphi'} \\ \text{where} \quad a' &= \frac{1}{(1 - \rho^2) a^\ast} = \frac{a}{|a|^2 - R^2}, \\ R' &= \frac{\rho }{|(1 - \rho^2) a^\ast|} = \frac{R}{||a|^2 - R^2|}, \\ e^{i \varphi'} &= - \frac{\rho + e^{- i \varphi}}{1 + \rho e^{- i \varphi}} \\ |e^{i \varphi'}|^2 &= \frac{\rho + e^{- i \varphi}}{1 + \rho e^{- i \varphi}} \frac{\rho + e^{i \varphi}}{1 + \rho e^{ i \varphi}} = \frac{\rho^2 + 1 + 2 \rho \cos \varphi }{1 + \rho^2 + 2 \rho \cos \varphi} = 1 \end{aligned}$ Therefore $\varphi' \in \mathbb{R}$ and all points of the image are one a circle with center $a'$ and radius $R'$ . Since the map $f$ is bijective and continuous, the circle must also be completely traversed.

A straight line that does not go through the origin can be represented in the complex plane by the points $z = u + i u \lambda, \quad \lambda \in \mathbb{R}$ where $u$ is the distance vector from the origin to the line and $i u$ is a direction vector that is parallel to the line. The image of this line then gives $\begin{aligned} f(z) &= \frac{1}{u^\ast - i u^\ast \lambda} \\ &= \frac{1}{u^\ast - i u^\ast \lambda} \frac{u + i u \lambda}{u + i u \lambda} \\ &= \frac{u + i u \lambda}{|u|^2 ( 1 + \lambda^2)} \\ &= \frac{u}{2|u|^2} \left(\frac{2}{1 + \lambda^2} + i \frac{2\lambda}{1 + \lambda^2} \right) \\ &= \frac{u}{2|u|^2} \left(1 - 1 + \frac{2}{1 + \lambda^2} + i \frac{2\lambda}{1 + \lambda^2} \right) \\ &= \frac{u}{2|u|^2} \Bigg(1 + \underbrace{\frac{1 - \lambda^2}{1 + \lambda^2}}_{v} + i \underbrace{\frac{2\lambda}{1 + \lambda^2}}_{w} \Bigg) \\ &= u' \left(1 + e^{i \varphi} \right)\\ \text{where} \quad u' &= \frac{u}{2|u|^2}, \\ \varphi &= \arctan \frac{w}{v} \\ v^2 + w^2 &= \frac{(\lambda^4 - 2 \lambda^2 + 1) + 4 \lambda^2 }{\lambda^4 + 2 \lambda^2 + 1} = 1 \end{aligned}$ Thus, all points lie on a circle around the point $u '$ with the radius $| u' |$ . This circle goes in particular through the origin $O$ . However, the origin itself is reached only in the limit case $\lambda \to \pm \infty$ , since $\lim_{\lambda \to \pm \infty} v = -1$ und $\lim_{\lambda \to \pm \infty} u = \pm 0$ . QED

More information about this so-called circle inversion can be found on Wikipedia.