Question

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

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

Answer 1

The correct option is B

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

Explanation for correct option

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

$\therefore z\overline{z}+a\overline{z}+\overline{a}z+b=0$

Let $z=x+iy$ and $a=\alpha +i\beta$

$$\begin{gathered} \Rightarrow \left(x+iy\right)\left(x-iy\right)+\left(\alpha +i\beta \right)\left(x-iy\right)+\left(\alpha -i\beta \right)\left(x+iy\right)+b=0 \\ \Rightarrow x^2+y^2+2\left(\alpha x+\beta y\right)+b=0 \end{gathered}$$

We know for the circle $x^2+y^2+2gx+2fy+c=0$ the radius is $R=\sqrt{g^2+f^2-c}$

So comparing with the standard form we have

$$\begin{gathered} R=\sqrt{{\alpha }^2+{\beta }^2-b} \\ \because R\ge 0 \\ \Rightarrow {\alpha }^2+{\beta }^2-b\ge 0 \\ \Rightarrow {\alpha }^2+{\beta }^2\ge b \\ \Rightarrow {\left|a\right|}^2\ge b\left[\because a\overline{a}={\left|a\right|}^2\right] \end{gathered}$$

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