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JEE Main 202529 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2025 · 29 January · Morning Shift · Mathematics · Complex Number

Let (|z_1-8-2 i| ≤ 1) and (|z_2-2+6 i| ≤ 2, z_1, z_2 ∈ C). Then the...

Let (\left|z_1-8-2 i\right| \leq 1) and (\left|z_2-2+6 i\right| \leq 2, z_1, z_2 \in \mathbf{C}). Then the minimum value of (\left|z_1-z_2\right|) is :Let (|z_1-8-2 i| ≤ 1) and (|z_2-2+6 i| ≤ 2, z_1, z_2 ∈ C). Then the minimum value of (|z_1-z_2|) is :

Options

  1. A13
  2. B10
  3. C3
  4. D7

Correct answer

D. 7

Step-by-step solution

Let (|z_1-8-2 i| ≤ 1) and (|z_2-2+6 i| ≤ 2, z_1, z_2 ∈ C). Then the minimum value of (|z_1-z_2|) is :
∵AB=100=10∴∣Z1−Z2∣min⁡=10−2−1=7\begin{aligned} & \because \mathrm{AB}=\sqrt{100}=10 \\ & \therefore\left|\mathrm{Z}_1-\mathrm{Z}_2\right|_{\min }=10-2-1=7\end{aligned}
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