I'm trying to prove that the set of $z$ that satisfies the following equation represents a circle or a straight line on $\mathbb C$.
$(a\bar c-c\bar a)|w^2|+(a\bar d-c\bar b)w+(b\bar c-d\bar a)\bar w+b\bar d-d\bar b=0 $ If $(a\bar c-c\bar a)=0$, I could prove that this is a straight line. (It took very long though. If you know how to show it simply, please give me some ideas on how to proceed.)
Apparently, if $(a\bar c-c\bar a)\neq 0$, we can get $\Bigg |\displaystyle w+\frac{\bar a d-\bar c b}{\bar a c-\bar c a}\Bigg|=\Bigg |\frac{ad-bc}{\bar a c-\bar c a}\Bigg |.$But I don't know how it works. I tried to factorize the first equation, but it didn't work. So I tried to expand the second equation by replacing the absolute value with the conjugate, but it didn't give me any good idea. Could someone give me some clues?