Question

The equation ${\left|z-z_1\right|}^2+{\left|z-z_2\right|}^2=k,k\in R$ represents a circle if

• A. $k\ge \frac{1}{2}{\left|z_1-z_2\right|}^2$
• B. $k\le \frac{1}{2}{\left|z_1-z_2\right|}^2$
• C. $k\ge \frac{1}{2}{\left|z_1+z_2\right|}^2$
• D. $k\le \frac{1}{2}{\left|z_1+z_2\right|}^2$

Answer 1

Answer: (A)

We have, ${\left|z-z_1\right|}^2+{\left|z-z_2\right|}^2=k$
$\Rightarrow 2|z{|}^2+{\left|z_1\right|}^2+{\left|z_2\right|}^2-2\text{Re}\left(z{\bar{z}}_1\right)-2\text{Re}\left(z{\bar{z}}_2\right)=k$
$\Rightarrow 2|z{|}^2-2\text{Re}\left\{z\left({\bar{z}}_1+{\bar{z}}_2\right)\right\}=k-\left({\left|z_1\right|}^2+{\left|z_2\right|}^2\right)$
$\Rightarrow |z{|}^2-\text{Re}\left\{z\left({\bar{z}}_1+{\bar{z}}_2\right)\right\}=\frac{1}{2}\left(k-{\left|z_1\right|}^2-{\left|z_2\right|}^2\right)$
$\Rightarrow {\left|z-\frac{z_1+z_2}{2}\right|}^2-\frac{1}{4}{\left|z_1+z_2\right|}^2=\frac{1}{2}\left(k-{\left|z_1\right|}^2-{\left|z_2\right|}^2\right)$
$\Rightarrow {\left|z-\frac{z_1+z_2}{2}\right|}^2=\frac{1}{2}k-\frac{1}{4}\left\{{\left|z_1\right|}^2+{\left|z_2\right|}^2-2\text{Re}\left(z_1{\bar{z}}_2\right)\right\}$
$=\frac{1}{2}k-\frac{1}{4}{\left|z_1-z_2\right|}^2$
$\Rightarrow {\left|z-\frac{z_1+z_2}{2}\right|}^2=\frac{1}{4}\left(2k-{\left|z_1-z_2\right|}^2\right)$
which will represent a real circle having centre at $\frac{z_1+z_2}{2}$ and radius $=\frac{1}{2}\sqrt{2k-{\left|z_1-z_2\right|}^2}$,
provided $k\ge \frac{1}{2}{\left|z_1-z_2\right|}^2$