Möbius transformations lines and circles I am looking for a basic outline of a proof
I know that all MT's are of the form $\frac{ax+b}{cx+d}$
For $c=0$, I know that lines/circles are preserved because translations and dilations do not change a line/circle from being a line/circle
But I am not sure how to prove this for all cases
My exam is actually tomorrow, so it would be great if someone could help me today :)
 A: Note that
$$
\dfrac{a\color{#C00000}{z}+b}{c\color{#C00000}{z}+d}=\frac ac+\frac1c\frac{bc-ad}{c\color{#C00000}{z}+d}\tag{1}
$$
is a composition of translation, dilation, and inversion $\left(z\mapsto\frac1z\right)$. Geometrically, it is easy to see that translation and dilation preserve circles and lines.

Inversion Preserves Circles Not Passing Through The Origin
If $|z-c|=r$ where $|c|\ne r$, then $\color{#C00000}{r^2}=(z-c)(\bar{z}-\bar{c})=\color{#C00000}{|z|^2+|c|^2-2\mathrm{Re}(z\bar{c})}$ and
$$
\begin{align}
&\left[\frac1z-\frac1c\left(\frac{|c|^2}{|c|^2-r^2}\right)\right]
\left[\frac1{\bar{z}}-\frac1{\bar{c}}\left(\frac{|c|^2}{|c|^2-r^2}\right)\right]\\
&=\frac1{|z|^2}+\frac1{|c|^2}\left(\frac{|c|^2}{|c|^2-r^2}\right)^2-2\mathrm{Re}\left(\frac1{z\bar{c}}\right)\left(\frac{|c|^2}{|c|^2-r^2}\right)\\
&=\frac1{|z|^2}+\frac1{|c|^2}\left(\frac{|c|^2}{|c|^2-r^2}\right)^2-\frac{\color{#C00000}{2\mathrm{Re}(z\bar{c})}}{|z|^2|c|^2}\left(\frac{|c|^2}{|c|^2-r^2}\right)\\
&=\frac1{|z|^2}+|c|^2\left(\frac1{|c|^2-r^2}\right)^2-\frac{\color{#C00000}{|z|^2+|c|^2-r^2}}{|z|^2}\left(\frac1{|c|^2-r^2}\right)\\
&=\left(\frac{r}{|c|^2-r^2}\right)^2\tag{2}
\end{align}
$$
Therefore, $\frac1z$ lies on the circle with
$$
\text{center}=\frac{\bar{c}}{|c|^2-r^2}\qquad\text{and}\qquad\text{radius}=\frac{r}{|c|^2-r^2}\tag{3}
$$

Limiting To Circles Passing Through The Origin
If we fix $c$ and let $r\to|c|$, the radius tends to $\infty$ while containing the following point:
$$
\begin{align}
\frac{\bar{c}}{|c|^2-r^2}-\frac{\bar{c}}{|c|}\frac{r}{|c|^2-r^2}
&=\frac{\bar{c}}{|c|}\frac{|c|-r}{|c|^2-r^2}\\
&=\frac{\bar{c}}{|c|}\frac1{|c|+r}\\
&\to\frac1{2c}\tag{4}
\end{align}
$$
Thus, the circle tends to a line passing through $\dfrac1{2c}$ and perpendicular to $\bar{c}$.

Lines
Since the inverse of $z\mapsto\frac1z$ is itself, we can invert the previous section to yield that a line maps to a circle passing through the origin whose center is $\dfrac1{2p}$, where $p$ is the closest point on the line to the origin.

Circles Whose Center Is The Origin And Lines Passing Through The Origin
In the previous sections, we have ignored the problems that occur when $c=0$ and $p=0$. These cases are trivial geometrically.
A: To begin with, if $c\neq0,$ then put $f_1(z)=z+d/c,$ $f_2(z)=1/z,$ $f_3(z)=\frac{bc-ad}{c^2}z,$ and $f_4(z)=z+a/c.$ Then $$(f_4\circ f_3\circ f_2\circ f_1)(z)=\frac{az+b}{cz+d}.$$ So, it suffices to show that each of $f_1,f_2,f_3,f_4$ maps generalized circles to generalized circles. The fact that $ad-bc\ne0$ will be essential here. The only somewhat tricky part is showing that this is true for $f_2.$ The idea is to show that lines are mapped to generalized circles through the origin and vice versa, while circles that don't pass through the origin are again mapped to circles that don't pass through the origin.
The $c=0$ case is straightforward.
