If $|z-2+i| \leq 2$, then the difference between the greatest and least value of $|z|$ is…
If $|z-2+i| \leq 2$, then the difference between the greatest and least value of $|z|$ is $(\mathrm{i}=\sqrt{-1})$
- $2 \sqrt{5}+4$
- $2 \sqrt{5}$
- $4$
- $8$
Solution
Note that $\left|z_1-z_2\right| \geq|| z_1|-| z_2||$
$\begin{array}{ll}
\therefore \quad & |z-2+i| \leq 2 \\
& \Rightarrow|| z|-| 2-i|| \leq 2 \\
& \Rightarrow-2 \leq|z|-|2-i| \leq 2
\end{array}$
$\begin{aligned}
& \Rightarrow-2 \leq|z|-\sqrt{4+1} \leq 2 \\
& \Rightarrow-2 \leq|z|-\sqrt{5} \leq 2 \\
& \Rightarrow \sqrt{5}-2 \leq|z| \leq 2+\sqrt{5}
\end{aligned}$
$\Rightarrow$ Largest value of $|z|$ is ' $2+\sqrt{5}$ ' and the least value is ' $\sqrt{5}-2$ '
$\therefore \quad$ Required difference $=2+\sqrt{5}-(\sqrt{5}-2)=4$
Asked in: MHT CET 2023 (12 May Shift 2)
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