Question

If z is a complex number, then the square of the radius of the circle $z\overline{z}-2(1+i)z-2(1-i)\overline{z}-1=0$ is

• A. $0$
• B. $1$
• C. $9$
• D. $4$

Answer 1

The correct option is C $9$

Equation of a circle with centre $z_0$ & radius r in its standard form is ,
$z\overline{z}-z_0\overline{z}-\overline{z_0}z+z_0\overline{z_0}-r^2=0$
comparing with the given equation,
$z\overline{z}-(-2-2i)z-(-2+2i)\overline{z}-1=0$
$\therefore 1=(r^2-z_0\overline{z_0})$
$\Rightarrow r^2=1+z_0\overline{z_0}$
$=1+{\left|z_0\right|}^2$
$=1+(\sqrt{(-2{)}^2+2^2}{)}^2$
$=1+(\sqrt{8}{)}^2$
$=1+8$
$=9$
$\therefore$ square of radius$=9$