Question

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

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

Answer 1

Answer: (B)

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

Adding and subtracting $a\overline{a}$ on both sides, the given relation becomes $(z+a)(\overline{z}+\overline{a})+b=a\overline{a}$

$(z+a)\left(\overline{z+a}\right)={\left|a\right|}^2-b$

${|z+a|}^2={\left|a\right|}^2-b$

The above equation represents a circle centred at $-a$ provided ${\left|a\right|}^2-b>0$ $($i.e., $\left|a^2\right|>b)$ as its radius is $\sqrt{{\left|a\right|}^2-b}.$

Note: If ${\left|a\right|}^2=b$ then it will be a point circle as its radius will be zero.