Question

If $z\bar{z}:+(3-4i)z+(3+4i)\bar{z}=0$ represent a circle, the 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)

Given $z\bar{z}+(3-4i)z+(3+4i)\bar{z}=0$
Let $z=x+iy$
$\therefore z\bar{z}=x^2+y^2$
$\Rightarrow x^2+y^2+(3-4i)(x+iy)+(3+4i)(x-iy)=0$
$\Rightarrow x^2+y^2+6x+8y=0$
$\Rightarrow (x^2+6x)+(y^2+8y)=0$
$\Rightarrow (x+3{)}^2+(y+4{)}^2=3^2+4^2$
$\Rightarrow [x-(-3){]}^2+[y-(-4){]}^2=5^2$
$(\because R=$ radius $=5)$
So area of circle be $\pi R^2=25\pi$