Explanation

(b)
Let $\mathcal{z =}x + iy,$ therefore given equation becomes
$(x + iy)(x - iy) + (2 - 3i)(x + iy) + (2 + 3i)(x - iy) + 4 = 0$
$\Rightarrow x^{2} + y^{2} + 2x + 3y - 3ix + 2iy + 2x - 2iy + 3ix + 3y + 4 = 0$
$\Rightarrow x^{2} + y^{2} + 4x + 6y + 4 = 0$
Therefore, given equation represents a circle with radius
$= \sqrt{2^{2} + 3^{2} - 4}$
$= \sqrt{4 + 9 - 4} = \sqrt{9} = 3$