= Solution
Use $z\overline z=|z|^2$ and $\overline Bz+B\overline z=2\operatorname{Re}(\overline Bz)$. When $A\ne0$, <completing the square> gives
$$
A|z|^2+\overline Bz+B\overline z+C=A\left|z+\frac BA\right|^2+C-\frac{|B|^2}{A}.
$$
Thus, with $\Delta=|B|^2-AC$,
$$
\boxed{\left|z+\frac BA\right|^2=\frac{\Delta}{A^2}\quad(A\ne0).}
$$
If $\Delta>0$, this is a <circle> with centre $-B/A$ and <radius> $\sqrt\Delta/|A|$. If $\Delta=0$, it is the single point $-B/A$, sometimes called a degenerate <circle> of <radius> zero. If $\Delta<0$, it is the <empty set>, because a squared <complex modulus> cannot be negative.
When $A=0$ and $B\ne0$, write $B=u+iv$ and $z=x+iy$. The remaining equation is
$$
\boxed{ux+vy=-\frac C2,\qquad (u,v)\ne(0,0),}
$$
which is a <straight line> with <normal vector> $(u,v)$. It passes through zero exactly when $C=0$. Finally, when $A=B=0$, the equation has no solutions if $C\ne0$ and every <complex number> is a solution if $C=0$.
The complete <classification of a real complex-circle equation> therefore includes two degeneracies in addition to ordinary <circles>, <straight lines> and the <empty set>:
$$
\boxed{\begin{array}{c|c}
\text{condition}&\text{locus}\\\hline
A\ne0,\ \Delta>0&\text{circle}\ (-B/A,\sqrt\Delta/|A|)\\
A\ne0,\ \Delta=0&\{-B/A\}\\
A\ne0,\ \Delta<0&\varnothing\\
A=0,\ B\ne0&\text{line }2\operatorname{Re}(\overline Bz)+C=0\\
A=B=0,\ C\ne0&\varnothing\\
A=B=C=0&\mathbb C
\end{array}}
$$
For the three advertised alternatives to be exhaustive, one must exclude the identically zero equation and either exclude or explicitly allow <radius>-zero <circles>. This is a qualification of the unrestricted coefficients, not a reason to discard the genuine <circle> and <straight line> cases used next.
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