Showing that $s|z|^2+cz+\bar{c}\bar{z}+t=0$ ($s\ne 0$) is always a circle in the complex plane

Question

I have used $f(z)=|z-z_0|-r=0$ to show that $s|z|^2+cz+\bar{c}\bar{z}+t=0$ ($s\ne 0$) ($s,t\in\mathbb{R},c\in\mathbb{C}$) is an equation for a circle, with $c = -s\bar{z}_0$ and $t=s|z_0|^2-sr^2$.

Now I want to go the other way: start from $s|z|^2+cz+\bar{c}\bar{z}+t=0$ and somehow deduce that this is an equation for a circle in $\mathbb{C}$. Can I just use the values for $c$ and $t$ that I have found earlier?

So, dividing by $s \ne 0$ and we get $|z|^2+\frac cs z+\frac{\bar{c}}s\bar{z}+\frac ts=0$ . Now choose $z_0 = -\frac{\bar{c}}{s}$ and $r^2 = |z_0|^2 - \frac ts$. And so on...

This seems very forced. Is there a less artificial way to proove this? Thanks.

Answer 1

It will be a circle only if $t$ and $s$ are real and such that $r^2=|z_0|^2-\frac{t}{s}>0.$