[UPLIFTJEE]

Let $\left|\mathrm{z}_{1}-8-2 \mathrm{i}\right| \leq 1$ and $\left|\mathrm{z}_{2}-2+6 \mathrm{i}\right| \leq 2, \mathrm{z}_{1}, \mathrm{z}_{2} \in \mathrm{C}$. Then the minimum value of $\left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|$ is [JEE (Main) 29 Jan 2025 S I]
(A) 13
(B) 3
(C) 7
(D) 10

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[UPLIFTJEE]

Solution

$\left|\mathrm{z}_{1}-(8+2 \mathrm{i})\right| \leq 1$
$\mathrm{z}_{1}$ is region inside \& circumference of circle,
centre $\mathrm{C}_{1}(8,2)$, radius $\mathrm{r}_{1}=1$
$\left|\mathrm{z}_{2}-2+6 \mathrm{i}\right| \leq 2, \mathrm{z}_{2}$ is region inside \& circumference of circle,
centre $\mathrm{C}_{2}(2,-6)$, radius $\mathrm{r}_{2}=2$
Now $\left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|_{\text {min }}=\mathrm{PQ}=\mathrm{C}_{1} \mathrm{C}_{2}-\mathrm{r}_{1}-\mathrm{r}_{2}$

\[ =\sqrt{6^{2}+8^{2}}-1-2=7 \]

[UPLIFTJEE]

Answer

C

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