Let $K\in\mathbb{R}$ and $z_1,z_2\in\mathbb{C}$.
Prove that the equation $|z-z_1|^2+|z-z_2|^2=K$ represent a circle iff $K\geq\frac{1}{2}|z_1-z_2|^2$.
My Attempt $$ |z|^2+|z_1|^2-2\mathcal{Re}(z\bar{z}_1)+|z|^2+|z_2|^2-2\mathcal{Re}(z\bar{z}_2)=K\\ 2|z|^2+|z_1|^2+|z_2|^2-2\bigg[\mathcal{Re}(z\bar{z}_1)+\mathcal{Re}(z\bar{z}_2)\bigg]=K\\ 2|z|^2+|z_1|^2+|z_2|^2-2\bigg[ a_1x+b_1y+a_2x+b_2y \bigg]=K\\ 2|z|^2+|z_1|^2+|z_2|^2-2\mathcal{Re}\bigg[ z\bar{z}_1+z\bar{z}_2 \bigg]=K\\ 2|z|^2+|z_1|^2+|z_2|^2-2\mathcal{Re}\bigg[ z(\bar{z}_1+\bar{z}_2) \bigg]=K\\ |z|^2+\frac{1}{2}(|z_1|^2+|z_2|^2)-2\mathcal{Re}\Big(z.\frac{\bar{z}_1+\bar{z}_2}{2}\Big)=\frac{K}{2}\\ |z|^2+\frac{1}{2}(|z_1|^2+|z_2|^2)+\Big|\frac{z_1+z_2}{2}\Big|^2-2\mathcal{Re}\Big(z.\frac{\bar{z}_1+\bar{z}_2}{2}\Big)=\frac{K}{2}+\Big|\frac{z_1+z_2}{2}\Big|^2\\ |z|^2+\big|\frac{z_1+z_2}{2}\big|^2-2\mathcal{Re}\Big(z.\frac{\bar{z}_1+\bar{z}_2}{2}\Big)=\frac{K}{2}+\Big|\frac{z_1+z_2}{2}\Big|^2-\frac{1}{2}(|z_1|^2+|z_2|^2)\\ \Big|z-\frac{z_1+z_2}{2}\Big|^2=\frac{K}{2}+\frac{1}{2}(z_1\bar{z}_2+\bar{z}_1z_2)=\frac{K}{2}+\frac{|z_1|^2+|z_2|^2}{4}+\frac{z_1\bar{z}_2+\bar{z}_1z_2}{4}-\frac{1}{2}(z_1\bar{z}_2+\bar{z}_1z_2)\\ \Big|z-\frac{z_1+z_2}{2}\Big|^2=\frac{K}{2}+\frac{1}{4}(|z_1|^2+|z_2|^2-z_1\bar{z}_2-\bar{z}_1z_2)=\frac{K}{2}-\frac{1}{4}|z_1-z_2|^2\\ \Big|z-\frac{z_1+z_2}{2}\Big|^2=\frac{1}{4}\Big[2K-|z_1-z_2|^2\Big]\\ \implies 2K\geq|z_1-z_2|^2\implies \boxed{K\geq\frac{1}{2}|z_1-z_2|^2} $$
As it was asked as a multiple choice question I think my attempt is a bit long and cumbersome. So, is there an easier way to solve this ?
So if point $T$ is for $z$ and $A,B$ for $z_1,z_2$ then we have
$$ AT^2+BT^2=K$$
Let $M = (A+B)/2$ be a midpoint for $AB$ then we have: $$ K= (T-A)^2+(T-B)^2$$ $$ = 2T^2-2T(A+B)+A^2+B^2 $$ $$ = 2T^2-4T\cdot M + 2M^2- 2M^2+A^2+B^2$$
$$ = 2(T-M)^2+(A-B)^2/2$$
So $$ (T-M)^2= K-(A-B)^2/4$$
and thus $4K \geq (A-B)^2$ or c) on your answer sheet.