Question

The equation $z\overline{z}+a\overline{z}+\overline{a}z+b=0,b\epsilon R$ represents a real circle with non zero radius if

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

Answer 1

Answer: (B)

$|a{|}^2>b$

Given that, the equation $z\overline{z}+a\overline{z}+\overline{a}z+b=0,b\epsilon R$ represents a real circle with non zero radius.
Let $a=\alpha +i\beta$ and $z=x+iy$
$\Rightarrow (x+iy)(x-iy)+(\alpha +i\beta )(x-iy)+(\alpha -i\beta )(x+iy)+b=0$
$\Rightarrow x^2+y^2+2\alpha x+2\beta y+b=0$
$r>0$
$\Rightarrow \sqrt{{\alpha }^2+{\beta }^2-b}>0$
$\Rightarrow {\alpha }^2+{\beta }^2>b$
$\therefore |a{|}^2>b$
Hence, option B.