Question

$|z-z_1{|}^2+|z-z_2{|}^2=a$ will represent a real circle on the argand plane if

• A. $a\ge |z_1-z_2{|}^2$
• B. $2a\ge |z_1-z_2{|}^2$
• C. $a\ge |z_1-z_2|$
• D. $2a\ge |z_1-z_2|$
  

Answer 1

The correct option is B $2a\ge |z_1-z_2{|}^2$

$|z-z_1{|}^2+|z-z_2{|}^2=a$
$\Rightarrow (z-z_1)(\overline{z}-{\overline{z}}_1)+(z-z_2)(\overline{z}-{\overline{z}}_2)=a$
$\Rightarrow 2z\overline{z}-z({\overline{z}}_1+{\overline{z}}_2)-\overline{z}(z_1+z_2)+z_1{\overline{z}}_1+z_2{\overline{z}}_2=a$
$\Rightarrow z\overline{z}-\left(\frac{{\overline{z}}_1+{\overline{z}}_2}{2}\right)z-\left(\frac{z_1+z_2}{2}\right)\overline{z}+\frac{z_1{\overline{z}}_1+z_2{\overline{z}}_2-a}{2}=\mathrm{0....}\left(1\right)$

Let's compare it with the general equation of the circle
$|z-\alpha |=r\Rightarrow |z-\alpha {|}^2=r^2\Rightarrow |z{|}^2-z\overline{\alpha }-\alpha \overline{z}+|\alpha {|}^2-r^2=\mathrm{0.....}\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$\Rightarrow \alpha =\frac{z_1+z_2}{2}\&|\alpha {|}^2-r^2=\frac{z_1{\overline{z}}_1+z_2{\overline{z}}_2-a}{2}$

Now $r^2\ge 0$
$\Rightarrow \frac{(z_1+z_2)({\overline{z}}_1+{\overline{z}}_2)}{4}\ge \frac{z_1{\overline{z}}_2+z_2{\overline{z}}_2-a}{2}$
$\Rightarrow z_1{\overline{z}}_1+z_1{\overline{z}}_2+{\overline{z}}_1{z}_2+z_2{\overline{z}}_2\ge 2(z_1{\overline{z}}_1+z_2{\overline{z}}_2)-2a$
$\Rightarrow 2a\ge z_1{\overline{z}}_1+z_2{\overline{z}}_2-z_1{\overline{z}}_2-{\overline{z}}_1{z}_2$
$\Rightarrow 2a\ge z_1({\overline{z}}_1-{\overline{z}}_2)-z_2({\overline{z}}_1-{\overline{z}}_2)$
$\Rightarrow 2a\ge (z_1-z_2)({\overline{z}}_1-{\overline{z}}_2)$
​​​​​​​$\Rightarrow 2a\ge (z_1-z_2)\left(\overline{z_1-z_2}\right)$
$\Rightarrow 2a\ge |z_1-z_2{|}^2$