Roots of $a z \overline{z} + \overline{b}z + b \overline{z}+c$: is my third case correct? Let $a,c \in \mathbb R$ and $b \in \mathbb C$ with $|b|^2 - ac > 0$.
To solve this I  distinguished 4 cases. Now I am interested in the case 

$a \neq0, c= 0$

-

Please can you tell me if my solution is correct?

If $a \neq 0$ and $c = 0$: 
Then $b \neq 0$ and the equation reduces to
$$ a z \overline{z} + \overline{b}z + b \overline{z} = 0$$ 
Divide by $a$ and complete the square by adding ${|b|^2 \over a}$ on both sides to get the equation
$$ z \overline{z} + {\overline{b} \over a}z + {b\over a} \overline{z} + {|b|^2 \over a} = (z - {b \over a})\overline{(z - {b\over a})} = |z - {b\over a}|^2 = {|b|^2 \over a}$$
The set of $z$ satisfying this equation is a circle of radius ${|b|\over \sqrt{a}}$ around ${b \over a}$. 
Since the radius has to be a real number this problem only has solutions if $a>0$.

It's “almost” fine. You can set $B=b/a$ and write
$$
z\bar{z}+\bar{B}z+B\bar{z}+B\bar{B}=B\bar{B}
$$
so
$$
(z+B)(\bar{z}+\bar{B})=|B|^2
$$
and this becomes
$$
|z+B|^2=|B|^2
$$
No condition on $a$ is necessary.
Another possibility is to write $B=h+ik$ and $z=x+iy$; then
$$
x^2+y^2+(h-ik)(x+iy)+(h+ik)(x-iy)=0
$$
that becomes
$$
x^2+y^2+2hx+2ky=0
$$
that's the equation of a circle.
