 Solution

Use $z\overline{z}=|z{|}^2$ and $\overline{B}z+B\overline{z}=2\mathrm{Re}(\overline{B}z)$. When $A\ne 0$, completing the square givesThus, with $\Delta =|B{|}^2-AC$,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 $B\ne 0$, write $B=u+iv$ and $z=x+iy$. The remaining equation iswhich 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 $C\ne 0$ 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: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.