Answer:
Option B
Explanation:
Consider the equation
$z\overline{z} + (2 -3i)z + (2+3i)\overline{z} + 4 = 0$ ....(1)
$Let z= x + iy and \overline{z} = x - iy, z\overline{z} = x^{2} + y^{2}$
Put values of z, $\overline{z}$ and $z\overline{z}$ in equation 1, we get
(x2 + y2) + (2 - 3i) (x + iy) + (2 + 3i) (x - iy) + 4 = 0
4x + 6y + 4 + x2 + y2 = 0
Now, we make it perfect square
x2 + y2 + 4x + 6y + 4 + 4 + 9 = 4+9
(x+2)2 + (y+3)2 = 9
This represents a circle of radius 3
