I want to show that a special conformal transformation maps line and circle to line and circle and find an example of it maps line to circle. The statement seems very true from the picture of the wiki page.
My thought is from here $$x'^\mu = \frac{x^\mu-b^\mu x^2}{1-2b\cdot x+b^2x^2}$$ I derive $$z'=\frac{z}{1+\bar b z}$$ And this seems to be an element of $\text{PSL}(2,\mathbb{C})$. And then I want to say an element of $\text{PSL}(2,\mathbb{C})$ maps line and circle to line and circle (which I don't know how to prove either).
Another way I tried is to start with a line equation say $x^2 = kx^1 + b$ and transform under the transformation law, and try to simplify it to a circle equation, but it looks very bad.
Shortly after I ask the question I find a simple solution.
First one the wiki page in the question, the SCT can be think of a as a composition of an inversion , a translation, and an inversion. It is clear the translation preserve line and circle. So we only need to show line and circle goes to line and circle under inversion.
Then we write down the circle equation in complex form: $$A z \bar z + B z + \bar B \bar z + C = 0$$ where $A, C\in \mathbb{R}$.
Notice when $A=0$ this is a line.
Now it is trivial to show under $z\rightarrow\frac{1}{z}$, this equation transform to $$C z \bar z + \bar B z + B \bar z + A = 0$$ which is the same form as before.
Therefore, we show a circle or line maps to circle or line under inversion, thus under SCT.