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If $|z-3+2 i| \leq 4$ then the difference between the greatest value and the least value of $|z|$ is
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Verified Answer
The correct answer is:
$2 \sqrt{13}$
$2 \sqrt{13}$
$|z-(3-2 i)| \leq 4$ represents a circle whose centre is $(3,-2)$ and radius $=4$.
$|z|=|z-0|$ represents the distance of point ' $z$ ' from origin $(0,0)$
Suppose $R S$ is the normal of the circle passing through origin ' $O$ ' and $G$ is its center $(3,-2)$
Here, $O R$ is the least distance and $O S$ is the greatest distance $O R=R G-O G$ and $O S=O G+G S \ldots(1)$ As, $R G=G S=4$ $O G=\sqrt{3^2+(-2)^2}=\sqrt{9+4}=\sqrt{13}$ From (1), $O R=4-\sqrt{13}$ and $O S=4+\sqrt{13}$ So, required difference
$$
\begin{aligned}
&=(4+\sqrt{13})-(4-\sqrt{13}) \\
&=\sqrt{13}+\sqrt{13}=2 \sqrt{13}
\end{aligned}
$$
$|z|=|z-0|$ represents the distance of point ' $z$ ' from origin $(0,0)$
Suppose $R S$ is the normal of the circle passing through origin ' $O$ ' and $G$ is its center $(3,-2)$
Here, $O R$ is the least distance and $O S$ is the greatest distance $O R=R G-O G$ and $O S=O G+G S \ldots(1)$ As, $R G=G S=4$ $O G=\sqrt{3^2+(-2)^2}=\sqrt{9+4}=\sqrt{13}$ From (1), $O R=4-\sqrt{13}$ and $O S=4+\sqrt{13}$ So, required difference
$$
\begin{aligned}
&=(4+\sqrt{13})-(4-\sqrt{13}) \\
&=\sqrt{13}+\sqrt{13}=2 \sqrt{13}
\end{aligned}
$$
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