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Q.
The equation $\left|z-z_{1}\right|^{2}+\left|z-z_{2}\right|^{2}=k, k \in R$ represents a circle if
Complex Numbers and Quadratic Equations
Solution:
We have, $\left|z-z_{1}\right|^{2}+\left|z-z_{2}\right|^{2}=k$
$\Rightarrow 2|z|^{2}+\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-2 \text{Re}\left(z \bar{z}_{1}\right)-2 \text{Re}\left(z \bar{z}_{2}\right)=k$
$\Rightarrow 2|z|^{2}-2 \text{Re}\left\{z\left(\bar{z}_{1}+\bar{z}_{2}\right)\right\}=k-\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)$
$\Rightarrow |z|^{2}-\text{Re}\left\{z\left(\bar{z}_{1}+\bar{z}_{2}\right)\right\}=\frac{1}{2}\left(k-\left|z_{1}\right|^{2}-\left|z_{2}\right|^{2}\right)$
$\Rightarrow \left|z-\frac{z_{1}+z_{2}}{2}\right|^{2}-\frac{1}{4}\left|z_{1}+z_{2}\right|^{2}=\frac{1}{2}\left(k-\left|z_{1}\right|^{2}-\left|z_{2}\right|^{2}\right)$
$\Rightarrow \left|z-\frac{z_{1}+z_{2}}{2}\right|^{2}=\frac{1}{2} k-\frac{1}{4}\left\{\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-2 \text{Re}\left(z_{1} \bar{z}_{2}\right)\right\}$
$=\frac{1}{2} k-\frac{1}{4}\left|z_{1}-z_{2}\right|^{2}$
$\Rightarrow \left|z-\frac{z_{1}+z_{2}}{2}\right|^{2}=\frac{1}{4}\left(2 k-\left|z_{1}-z_{2}\right|^{2}\right)$
which will represent a real circle having centre at $\frac{z_{1}+z_{2}}{2}$ and radius $=\frac{1}{2} \sqrt{2 k-\left|z_{1}-z_{2}\right|^{2}}$,
provided $k \geq \frac{1}{2}\left|z_{1}-z_{2}\right|^{2}$