Conformal circles (A property of inversions) I came across the following proposition in this book:

If C is a circle in the Euclidean plane, iC is conformal, that is it preserves angles. Also, iC takes circles not containing the center of C to circles, circles containing the center to lines, lines not containing the center to circles containing the center, and lines containing the center to themselves.

I could not figure out what this proposition means. I need to understand this statement before I move on to its proof.

A holomorphic mappingg $\mathbb{\hat{C}} \rightarrow \mathbb{\hat{C}}$ that has a holomorphic inverse is called a conformal mapping of $\mathbb{\hat{C}}$. It turns out that every conformal mapping of $\mathbb{\hat{C}}$ is a Möbius transformation.
The proposition you ask about asserts that a conformal map will send the points along a circle to either a circle or a line (depending on whether or not $0$ is on the circle) and sends the points on a line to lines or circle (depending on whether or not $0$ is on the circle). I give a relatively easy to follow proof of this below.
Theorem Any Möbius transformation is a composition of translations, dilation and inversions. (A Möbius transformation is a map of the form $\phi(z) = \displaystyle\frac{az+b}{cz+d}$.) A translation is of the form $\phi(z) = z+c$, a dilation is of the form $\phi(z) = \lambda z$, and an inversion is of the form $\displaystyle \phi(z) = \frac{1}{z}$.
Proof Let $\phi$ be given. If $\phi(\infty)=\infty$, then we know that $\phi(z) = a z+b=(z→z+b)◦(z→a z)$. Suppose $\phi(∞)=c \neq ∞$. Then $\psi=(z→1/z)◦(z→z−c)◦φ$ is a Möbius transformation fixing $\infty$:
$$\infty \xrightarrow{\psi} c \xrightarrow{z-c} 0 \xrightarrow{1/z} \infty$$
Thus, $ψ$ is a composition of translations, dilations and inversions. The same is true for $φ$.
Now that there is some background on the conformal maps we move to the actual problem at hand.
Theorem Möbius transformations preserve the class of circles and lines.
Proof Circles and lines are clearly preserved by translations and dilations.  It suffices to check that they are preserved under inversion. Consider a circle $C=\{z∈\mathbb{C}||z−a|=r\},\ r>0$. Put $w=1/z$.
\begin{align*}
z∈C&⇔|z−a|^2=r^2\\
&⇔|z|^2−\overline{a} z−a\overline{z}+|a|^2=r^2
\end{align*}
If|a|=r (equivalently, C goes through $0$), then
\begin{align*}
z∈C&⇔|z|^2−\overline{a} z−a\overline{z}=0\\
&⇔1−\overline{a w}−a w=0\ \text{or}\ w=∞\\
&⇔2Re(a w)=1\ \text{or}\ w=∞\\
&⇔Re(e^{iarg(a)}w)=\frac{1}{2|a|}\ \text{or}\ w=∞
\end{align*}
This last condition describes a line.
If $|a|\neq r$(equivalently, C does not go through 0), then
\begin{align*}
z∈C&⇔|z|^2−\overline{a} z−a\overline{z}+|a|^2−r^2=0\\
&⇔1−\overline{a w}−a w+(|a|^2−r^2)|w|^2=0\\
&⇔|w|^2−\frac{\overline{a w}+a w}{|a|^2−r^2}+\frac{1}{|a|^2−r^2}=0\\
&⇔|w|^2−\frac{\overline{a w}+aw}{|a|^2−r^2}+\frac{|a|^2}{(|a|^2−r^2)^2}+\frac{1}{|a|^2−r^2}−\frac{|a|^2}{(|a|^2−r^2)^2}=0\\
&⇔\left|w-\frac{|a|^2}{|a|^2−r^2}\right|^2−\frac{r^2}{(|a|^2−r^2)^2}=0
\end{align*}
The last equation describes a circle. A similar calculation shows that under inversion, the image of a line $\ell$ is a circle (if $0∉ \ell$) or a line (if $0∈\ell$).
