The equation of a circle on a complex plane?

Question

The equation of a circle $|z-z_0|=r$ in a complex plane has (among others) the form:

$$z\overline{z}+\overline{b}z+b\overline{z}+c=0$$ where $b=-z_0 \in \mathbb{C}$.

What I'd like to understand is, why is it so?

Answer 1

It's because you can write the original form as $$|z-z_0|^2=r^2$$ and $$|z-z_0|^2=(z-z_0)(\overline{z-z_0})=(z-z_0)(\overline{z}-\overline{z_0})$$ Now substitute for $z_0$