given equation represents the circle with center (3,-2) and is of radius $( R )=4$

$| z |$ represents the distance of point 'z' from origin

Greatest and least distances occur along the normal through the origin Normal always passes through center of circle

From figure; let PQ be the normal through origin 'O' and $C$ be its center (3,-2)

it is clear that OP is the least distance

and $OQ$ is the greatest distance

From diagram; $OP = CP – OC$ and $OQ = CQ + OC$

Here, $CP = CQ = R =4$

$OC =\sqrt{(3-0)^{2}+(-2-0)^{2}}$

$\Rightarrow OC =\sqrt{13}$

$\therefore OP = CP – OC$

$\Rightarrow OP =4-\sqrt{13}$

$\therefore$ Least distance $OP =4-\sqrt{13}$

and $\quad OQ = CQ + OC$

$\Rightarrow OQ =4+\sqrt{13}$

$\therefore$ Greatest distance $= OQ =4+\sqrt{13}$

Difference between greatest and least distance $=0 Q-O P=(4+\sqrt{13})-(4-\sqrt{13})$

$\Rightarrow$ Difference $=2 \sqrt{13}$

final answer=2 $\sqrt{13}$

the correct option is 'B'