Question

If $z\overline{z}+(3-4i)z+(3+4i)\overline{z}=0$ represent a circle then area of the circle in square units is

• A. $5\pi$
• B. $10\pi$
• C. $25{\pi }^2$
• D. $25\pi$

Answer 1

Answer: (D)

$25\pi$

Area of circle in standard from represented by complex number is,
$z\overline{z}-z_0\overline{z}-\overline{z_0}z+z_0\overline{z_0}-r^2=0$
where $z_0$: center of circle &
r: radius of circle
comparing with the giving equation.

$z_0=-(3+4i)$
$\therefore \overline{z_0}{z}_0={\left|z_0\right|}^2=25$

$\overline{z_0}{z}_0-r^2=0$
$\therefore 25-r^2=0$

$\therefore r^2=25$

Thus the area of the circle is $\pi r^2=25\pi$