The equation of a circle on a complex plane? 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?
 A: 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 $z_0$
A: A particularly simple equation is that of a circle:
$$\{z: |z - a| = r\}$$
is the circle with radius $r$ and center $a$. By squaring that equation we obtain
$$(z - a)(z' - a') = r²$$
or
$$zz' - (za' + z'a) + (aa' - r²) = 0$$
and finally
$$zz' - (za' + z'a) + s = 0$$
where $s$ is a real number. The circle is centered at a and has the radius $r = \sqrt{a}a' - s$, provided the root is real.
This representation of the circle is more convenient in some respects. For example, we may immediately check that the transformation $w = f(z) = \frac{1}{z}$ maps circles onto circles. Indeed, substituting $z = \frac{1}{w}$ we get
$$\frac{1}{w} \times \frac{1}{w'} - (\frac{a'}{w} + \frac{a}{w'}) + s = 0$$
which, if multiplied by $ww'$, leads to
$$ww' - (wb' + w'b) + t = 0$$
where $b = \frac{a'}{s}$ and $t = \frac{1}{s}$, an equation in the same form.
Letting $a = \alpha + i\beta$ yields yet another form of essentially same equation:
$$zz' - \alpha(z + z') - i\beta(z - z') + s = 0$$
where $\alpha$ and $\beta$ are both real. Yet the most general form of the equation is this
$$Azz' + Bz + Cz' + D = 0$$
