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I am trying to show that if $A, C$ are real numbers with $|B|^2 - AC > 0$ for $B \in \mathbb{C}$ and $A \neq 0$ then $Az\overline{z} + Bz + \overline{Bz} + C = 0$ is the locus of a circle, and the converse, that if $Az\overline{z} + Bz + \overline{Bz} + C = 0$ is the locus of a circle, then $A$ and $C$ are real numbers, $|B|^2 - AC > 0$ and $A \neq 0$.

My idea was to take some $w \in \mathbb{C}$ and expand $|w - z|^2 = r^2 \Longleftrightarrow |w|^2 + |z|^2 - w\overline{z} - \overline{w}z - r^2 = 0$. Then multiplying by $A$ we get that $A|z^2| + (-A\overline{w})z + (-Aw)\overline{z} + A(|w|^2 - r^2) = 0$.

The first direction is quite easy as if $A, C$ are real numbers with the said assumptions and $A \neq 0$, then the chain of equivalences concludes the claim.

Question: I am currently stuck at concluding the converse, that if $Az\overline{z} + Bz + \overline{Bz} + C = 0$ is the locus of a circle, then $A \neq 0$ and $|B|^2 - AC > 0$, as I don't see from the chain of equivalences how taking some $z$ from the circle forces $A$ to be non-zero. It is true that for $B \equiv -A\overline{w}$, $A$ and $C$ has to be non-zero. But then, how do we know a priori that our choice of $B \equiv -A\overline{w}$ is right?

Cartesian Bear
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    If $A$ is 0 and $B$ is not $0$, you get the equation of a real affine line (Set $B = B_1+iB_2$ and $z=x+iy$ with $B_1,B_2,x,y$ real numbers) to see that. If $A$ and $B$ are null, you get the whole plane or the empty set according whether $C$ is null or not. If a is not zero, you may divide by $A$ and transform the equation you can make $|z+\overline{B}/A|^2$ appear... – Christophe Leuridan Jun 08 '22 at 19:32
  • How about working by contradiction? – dfnu Jun 08 '22 at 19:32
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    Check this: https://math.stackexchange.com/q/2413830/42969, or this: https://math.stackexchange.com/q/2091238/42969 – Martin R Jun 08 '22 at 19:40

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