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JEE Main 20262 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2026 · 2 April · Evening Shift · Mathematics · Complex Number

Let the circles C_1 : |z| = r and C_2 : |z - 3 - 4i| = 5 , z ∈ ℂ ,...

Let the circles C1:∣z∣=rC_1 : |z| = r and C2:∣z−3−4i∣=5C_2 : |z - 3 - 4i| = 5, z∈Cz \in \mathbb{C}, be such that C2C_2 lies within C1C_1. If z1z_1 moves on C1C_1, z2z_2 moves on C2C_2 and min⁡∣z1−z2∣=2\min |z_1 - z_2| = 2, then max⁡∣z1−z2∣\max |z_1 - z_2| is equal to:

Options

  1. A1212
  2. B1717
  3. C2222
  4. D2424

Correct answer

C. 2222

Step-by-step solution

The center of circle C1C_1 is O1(0,0)O_1(0,0) and its radius is r1=rr_1 = r.

The center of circle C2C_2 is O2(3,4)O_2(3,4) and its radius is r2=5r_2 = 5.

The distance between the centers of the two circles is d=32+42=5d = \sqrt{3^2 + 4^2} = 5.

Since C2C_2 lies entirely within C1C_1, the minimum distance between a point on C1C_1 and a point on C2C_2 is given by r1−(d+r2)r_1 - (d + r_2).

We are given that min⁡∣z1−z2∣=2\min |z_1 - z_2| = 2.

r−(5+5)=2⇒r−10=2⇒r=12r - (5 + 5) = 2 \Rightarrow r - 10 = 2 \Rightarrow r = 12

The maximum distance between a point on C1C_1 and a point on C2C_2 is given by r1+d+r2r_1 + d + r_2.

max⁡∣z1−z2∣=12+5+5=22\max |z_1 - z_2| = 12 + 5 + 5 = 22

Answer: 2222
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