Question

Prove that $|z-z_1{|}^2+|z-z_2{|}^2=k$ will represent a circle, if $|z_1-z_2{|}^2\le 2k$.

Answer 1

$|z-z_1{|}^2+|z-z_2{|}^2=k$

$⟹2|z{|}^2+|z_1{|}^2+|z_2{|}^2-2Re\left(z{\overline{z}}_2\right)-2Re\left(z{\overline{z}}_1\right)=k$

$⟹|z{|}^2-Re\left(z\right({\overline{z}}_1+{\overline{z}}_2\left)\right)=\frac{1}{2}(k-|z_1{|}^2-|z_2{|}^2)$

$⟹|z-\frac{z_1+z_2}{2}{|}^2-\frac{1}{4}|z_1+z_2{|}^2=\frac{1}{2}(k-|z_1{|}^2-|z_2{|}^2)$

$⟹|z-\frac{z_1+z_2}{2}{|}^2=\frac{1}{2}k-\frac{1}{4}(|z_1{|}^2+|z_2{|}^2-2Re(z_1\overline{z_2}\left)\right)$

$⟹|z-\frac{z_1+z_2}{2}{|}^2=\frac{1}{4}(2k-|z_1-z_2{|}^2)$ which represents the real circle having centre at $\frac{z_1+z_2}{2}$

And radius is $\frac{1}{2}\sqrt{2k-|z_1-z_2{|}^2}$

$⟹2k\ge |z_1-z_2{|}^2⟹|z_1-z_2{|}^2\le 2k$