deriving relation for circle in complex plane

Question

Prove that $|z-z_1|²+|z-z_2|²=k$ will represent a circle if $|z_1-z_2|²\leq2k$

I tried using the concept of family of circles, but it didn't help me

Answer 1

Expressing z as $x+yi, z_1=a_1+b_1i, z_2=a_2+b_2i$ so re writing our equation we get $(x-a_1)^2+(y-b_1)^2+(x-a_2)^2+(y-b_2)^2=k$ through simplifying we get that this a circle

Answer 2

$$ \begin{align} k = |z-z_1|^2+|z-z_2|^2 & =(z-z_1)(\bar z - \bar z_1) + (z-z_2)(\bar z - \bar z_2) \\ & = 2 |z|^2 + |z_1|^2 + |z_2|^2 - z(\bar z_1 + \bar z_2) - \bar z(z_1+z_2) \\ & = 2 \left(z - \frac{z_1+z_2}{2}\right)\left(\bar z - \frac{\bar z_1+ \bar z_2}{2}\right) -\frac{1}{2}(z_1+z_2)(\bar z_1+ \bar z_2) + |z_1|^2+|z_2|^2 \\ & = 2 \left|z - \frac{z_1+z_2}{2}\right|^2 + \frac{1}{2}\left|z_1-z_2\right|^2 \end{align} $$

For $\left|z_1-z_2\right|^2 \le 2k$ the equation represents a circle of radius $\frac{1}{2}k - \frac{1}{4} \left|z_1-z_2\right|^2$ centered at $\frac{z_1+z_2}{2}\,$:

$$\left|z - \frac{z_1+z_2}{2}\right|^2 = \frac{1}{2}k - \frac{1}{4}\left|z_1-z_2\right|^2$$

(The relation is a direct translation in complex numbers of the median length formula in a triangle.)