Question

The equation $z\overline{z}+a\overline{z}+\overline{a}z+b=0,b$ belongs to $R$ represents a circle if

• A. ${\left|a\right|}^2={\left|b\right|}^2$
• B. ${\left|a\right|}^2>\left|b\right|$
• C. ${\left|a\right|}^2<{\left|b\right|}^{}$
• D. None of these

Answer 1

The correct option is B

${\left|a\right|}^2>\left|b\right|$

Solve the given equation in a simpler form.

The given equation is $z\overline{z}+a\overline{z}+\overline{a}z+b=0$.

$\begin{array}{rcl}z\overline{z}+a\overline{z}+\overline{a}z+b& =& 0\\ z\overline{z}+a\overline{z}+\overline{a}z+a\overline{a}& =& a\overline{a}-b\\ \left(z+a\right)\left(\overline{a}+\overline{z}\right)& =& a\overline{a}-b\end{array}$

So, ${\left|z+a\right|}^2=\left|a^2-b\right|$ represents a circle only if ${\left|a\right|}^2>b$.

Hence, option $\left(B\right)$ is the correct answer.