If |z - 3 + 2i| leq 4,then the difference bet | vedclass

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(B) The given equation $|z - (3 - 2i)| \leq 4$ represents a disk with center $C(3, -2)$ and radius $R = 4$.$|z|$ represents the distance of a point $z

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$2\sqrt{13}$

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(B) The given equation $|z - (3 - 2i)| \leq 4$ represents a disk with center $C(3, -2)$ and radius $R = 4$.$|z|$ represents the distance of a point $z$ from the origin $O(0, 0)$.The greatest and least distances from the origin to the circle occur along the line passing through the origin and the center $C$.Let $OC$ be the distance from the origin to the center $C(3, -2)$.$OC = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$.The least distance of $|z|$ from the origin is $|OC - R| = |\sqrt{13} - 4|$. Since $\sqrt{13} The greatest distance of $|z|$ from the origin is $OC + R = \sqrt{13} + 4$.The difference between the greatest and least values is $(4 + \sqrt{13}) - (4 - \sqrt{13}) = 2\sqrt{13}$.

If $|z - 3 + 2i| \leq 4$,then the difference between the greatest value and the least value of $|z|$ is

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