Four complex numbers $z_1,z_2,z_3,z_4$ lie on a generalized circle if and only if they have a real cross ratio $[z_1,z_2,z_3,z_4]\in\mathbb{R}$ Let $[z_1,z_2,z_3,z_4]$ denote the cross ratio of the complex numbers $z_1,z_2,z_3,z_4\in \mathbb{C}$. Show that the distinct points $z_1,z_2,z_3,z_4\in\widehat{\mathbb{C}}$ lie on a generalized circle if and only if $[z_1,z_2,z_3,z_4]\in\mathbb{R}$.

I saw this statement claimed in this answer, and wondered how one would go about showing it. I know of some facts we could work with:


*

*First, the definition: A generalized circle in $\mathbb{C}$ is either a circle or a line in $\mathbb{C}$.

*I know that a Mobius transformation maps generalized circles to generalized circles.

*I know that if $f$ is a Mobius transformation and $z_j'=f(z_j)$, $j=1,2,3,4$, then $[z_1',z_2',z_3',z_4']=[z_1,z_2,z_3,z_4]$.


I have posted a potential proof as an answer below. Please let me know if you think it's complete. Thank you!
 A: We will show (1) if $z_1,z_2,z_3,z_4$ lie on a generalized circle, then $[z_1,z_2,z_3,z_4]\in\mathbb R$, and (2) if $[z_1,z_2,z_3,z_4]\in\mathbb R$, then $z_1,z_2,z_3,z_4$ lie on a generalized circle.


*

*First, suppose $z_1,z_2,z_3,z_4$ lie on a generalized circle. We know that, given three real numbers $x_1,x_2,x_3 \in \mathbb{R}$, there exists a Mobius transformation
\begin{equation*}
F=F_{x_1,x_2,x_3}^{-1} \circ F_{z_1,z_2,z_3}
\end{equation*}
such that $F(z_i)=x_i$, $i=1,2,3$. Also, since Mobius transformations map generalized circles to generalized circles, we know that $z_4$ is also mapped to some $x_4 \in \mathbb{R}$. Therefore, since $x_1,x_2,x_3,x_4 \in \mathbb{R}$, then their cross ratio $[x_1,x_2,x_3,x_4]=\frac{x_1-x_3}{x_2-x_3}\cdot\frac{x_2-x_4}{x_1-x_4} \in \mathbb{R}$. Therefore, since the cross ratio is invariant under Mobius transformation, we have
\begin{align*}
 [z_1,z_2,z_3,z_4]
 &=[F(z_1),F(z_2),F(z_3),F(z_4)] \\
 &=[x_1,x_2,x_3,x_4] \\
 &=\frac{x_1-x_3}{x_2-x_3}\cdot\frac{x_2-x_4}{x_1-x_4} \\
 &\in \mathbb{R}
 \end{align*}

*Next, suppose $[z_1,z_2,z_3,z_4] \in \mathbb{R}$. Let
\begin{equation*}
F=F_{x_1,x_2,x_3}^{-1} \circ F_{z_1,z_2,z_3}
\end{equation*}
be the Mobius transformation which sends $F(z_i)=x_i$, $i=1,2,3$, and let $z_4'=F(z_4)$. Since the cross ratio is invariant under Mobius transformation, we know that 
\begin{equation*}
[z_1,z_2,z_3,z_4]=[x_1,x_2,x_3,z_4']
\end{equation*}
Therefore the cross ratio $[x_1,x_2,x_3,z_4']$ is also a real number, which implies that $z_4'$ is a real number as well. Now consider $F^{-1}$, which is also a Mobius transformation. Since $x_1,x_2,x_3,z_4'$ lie on a generalized circle (the real line), and since a Mobius transformation maps generalized circles to generalized circles, then $z_1,z_2,z_3,z_4 \in F^{-1}(\mathbb{R})$ belong to a generalized circle.

A: First, show that every generalized circle can be sent by a Mobius transformation to the real line (transitivity). 
Then, all you have to do is to show that if the cross ratio of $4$ real points is always real. 
Also note that this theorem can be shown with elementary geometry using the inscribed angle theorem.
