I'm trying to prove the following represents a circle but I can't see how to reduce it down to the form $|z-z_0|=c$:
Let $r \in \mathbb{R}, r \not= 0, c \in \mathbb{R}, k \in \mathbb{C} : |k|^2>cr$ Prove $r|z|^2 - \bar{k}z - k \bar{z} + c = 0$ represents a circle