Question

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

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

Answer 1

The correct option is B $|a{|}^2>b$

By adding $a\overline{a}$ on both the sides of $z\overline{z}+a\overline{z}+\overline{a}z=-b$
we get, $(z+a)(\overline{z}+\overline{a})=a\overline{a}-b$
$\Rightarrow |z+a{|}^2=|a{|}^2-b,\because z\overline{z}=|z{|}^2$
This equation will represent a circle with centre $z=-a,if|a{|}^2-b>0,i.e.|a{|}^2>b$ since $|a{|}^2=b$ represents point circle only.