Question

For equation $|z-z_1{|}^2+|z-z_2{|}^2=a$ to represent a circle, which of the following is/are true?

• A. Centre of the circle is $\frac{(z_1+z_2)}{4}.$
• B. Centre of the circle is $\frac{(z_1+z_2)}{2}.$
• C. The circle is real, if $2a\ge |z_1-z_2{|}^2.$
• D. The circle is real, if $4a\ge |z_1-z_2{|}^2.$

Answer 1

Answer: (B), (C)

The circle is real, if $2a\ge |z_1-z_2{|}^2.$

$|z-z_1{|}^2+|z-z_2{|}^2=a$
$\Rightarrow z\overline{z}-\left(\frac{\overline{z_1}+\overline{z_2}}{2}\right)z-\left(\frac{z_1+z_2}{2}\right)\overline{z}+\frac{z_1\overline{z_1}+z_2\overline{z_2}-a}{2}=0\cdots \left(1\right)$
Equation $\left(1\right)$ is of the form of $z\overline{z}+\overline{a}z+a\overline{z}+r=0$.
Hence, centre $=\frac{(z_1+z_2)}{2}$.
Also, equation $\left(1\right)$ will represent a real circle, if $a\overline{a}-r>0$
$\Rightarrow |a{|}^2-r>0$
$\Rightarrow \frac{|z_1+z_2{|}^2}{4}\ge \frac{z_1\overline{z_1}+z_2\overline{z_2}-a}{2}$
$\Rightarrow z_1\overline{z_1}+z_1\overline{z_2}+\overline{z_1}{z}_2+z_2\overline{z_2}>2(z_1\overline{z_1}+z_2\overline{z_2})-2a$
$\Rightarrow 2a\ge z_1\overline{z_1}+z_2\overline{z_2}-z_1\overline{z_2}-z_2\overline{z_1}$
$\Rightarrow 2a\ge |z_1-z_2{|}^2$