Question

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

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

Answer 1

The correct option is A $k\ge \frac{1}{2}{|z_1-z_2|}^2$

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

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

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

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

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

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

which will represent a real circle having center at $\frac{z_1+z_2}{2}$ and radius $=\frac{1}{2}\sqrt{2k-{|z_1-z_2|}^2}$, provided $k\ge \frac{1}{2}{|z_1-z_2|}^2$