orientation of circle in complex plane

Question

How to use the formula cross ratio $${Im}\frac{\frac{z-z_1}{z-z_2}}{\frac{z_3-z_1}{z_3-z_2}}$$ cross ratio$<0 $ or cross ratio $>0$ to determine the orientation of the circle and the left or right side of the circle?

Answer 1

See the graphical example that helps to build a conjecture... and prove it.

Let $(C)$ be the circle circumscribed to triangle $z_1z_2z_3$.

We start from the criteria using cross-ratio :

$$\tag{1}z \in (C) \ \ \iff \ \ c=[z_1,z_2;z_3,z]:=\frac{\frac{z-z_1}{z-z_2}}{\frac{z_3-z_1}{z_3-z_2}} \in \mathbb{R}$$

The question is : how can be interpreted the sign of this real number ?

See figure below where the color convention is that points $z$ are colored green (resp. red) if $c>0$ (resp $c<0$). Visibly there is a connection with arcs delimited by chord $[z_1z_2]$ (and always this one).

Now that we know what to prove, here is the proof.

let us write (1) under the equivalent form: $\frac{z_1-z}{z_2-z}=a \frac{z_1-z_3}{z_2-z_3}$ with $a \in \mathbb{R}.$

Taking arguments on each side, we have :

$$\underbrace{\arg(z_1-z)-\arg(z_2-z)}_{\text{angle}(ZZ_1,ZZ_2)}=\arg(a)+\underbrace{\arg(z_1-z_3)-\arg(z_2-z_3)}_{\text{angle}(Z_3Z_1,Z_3Z_2)}$$

Thus, either $a>0$ then $\arg(a)=0$, or $a<0$, in which case $\arg(a)=\pi$ (or $180°$ if you are accustomed to degrees).

In particular, if $Z$ is on the green arc, $Z$ "sees" line segment with a same angle as $Z_3$ sees it (the equal angle characterization of an arc of circle)

On the contrary, if $Z$ is on the red arc, $Z$ "sees" line segment with an angle shifted by $+180°$ (or $-180°$ which amounts to the same) : on the given figure, in such a case, $\text{angle}(ZZ_1,ZZ_2)\approx-120°$ (note the minus sign) whereas $\text{angle}(Z_3Z_1,Z_3Z_2)\approx60°.$ ($60°=180°-120°$)

Remark: There is a previous question on math SE Cross Ratio is positive real if four points on a circle but no answer addresses the positivity issue...