Derivation of condition for general circle equation in complex plane It can be shown that any line or circle can be written in the general form
$$ Az\bar{z}+\bar{B}z+B\bar{z}+D=0 $$ 
where $A, D$ are real and $B, z$ are complex.
However, it can also be shown that for this equation to be a circle, it needs to satisfy $B\bar{B}-AD>0$. I'm not sure how to derive this condition, but my guess is that it has something to do with the complex discriminant. The above equation can probably be solved for $| z |$, and the generalized formula is probably something like
$|z|=\frac{-(?)\pm \sqrt{|B|^2-4AD}}{2A}$
Would appreciate your insight.
 A: Let us expand the other equation for a circle expressing that there exist a center $z_0$ and a radius $R>0$ such that:
$$|z-z_0|=R \ \ \iff \ \ |z-z_0|^2=R^2 \ \ \iff (z-z_0)\overline{(z-z_0)}=R^2$$
$$\tag{1}\iff \ \ z \bar z - \bar{z_0}z - z_0\bar{z} + (|z_0|^2-R^2)=0$$
Identifying (1) with your expression, written under the form:
$$z\bar{z}+\frac{\bar{B}}{A}z+\frac{B}{A}\bar{z}+\frac{D}{A}=0$$
we get, in particular 
$$\tag{2}z_0=-\dfrac{B}{A}$$
and
$$|z_0|^2-R^2=\frac{D}{A}$$
$$\tag{3}\implies \ \ |z_0|^2-\frac{D}{A} = R^2 > 0.$$
Plugging into (3) the expression of $z_0$ obtained in (2) gives
$$\dfrac{B\bar{B}}{A^2}-\frac{D}{A} > 0 \ \ \iff \ \ B\bar{B}-AD>0.$$
A: If $A=0$ the equation reduces to $\;2 \operatorname{Re}(\bar B z)+D=0\;$ which is a line in the complex plane.
Otherwise divide by $A \ne 0$ and with $b = \frac{B}{A}, D=\frac{D}{A}$ the equation becomes:
$$ z\bar{z}+\bar{b}z+b\bar{z}+d=0 \;\iff\; (z+b)(\bar z + \bar b) - b \bar b + d = 0 \;\iff\; |z+b|^2 = |b|^2 -d$$ 
The latter has solutions iff the RHS is non-negative $\;|b|^2 -d \ge 0\;$ and the solution set is a proper circle of radius $\;\sqrt{|b|^2 -d}\;$ centered at $\;-b\;$ iff the RHS is strictly positive $\;|b|^2 -d \gt 0\,$.
A: Hint: Plug in $z = x+iy$ and complete the square. 
