JEE Main 2026 — Apr 2, Shift 2 — Mathematics solution

JEE Main2026Apr 2, Shift 2Mathematics

Let the circle $C_{1} : ∣ z ∣ = r$ and $C_{2} : ∣ z - 3 - 4 i ∣ = 5$ , $z \in C$ , be such that $C_{2}$ lies within $C_{1}$ . If $z_{1}$ moves on $C_{1}$ , $z_{2}$ moves on $C_{2}$ and $min ∣ z_{1} - z_{2} ∣ = 2$ , then $max ∣ z_{1} - z_{2} ∣$ is equal to :-

A
$12$
B
$17$
C
$22$
D
$24$

Solution

The centre of $C_{2}$ is at distance $3^{2} + 4^{2} = 5$ from the origin, which is also the centre of $C_{1}$ . Since $C_{2}$ lies inside $C_{1}$ , every point of $C_{2}$ is within distance $5 + 5 = 10$ of the origin.

For $z_{1}$ on $C_{1}$ (radius $r$ ) and $z_{2}$ on $C_{2}$ , the minimum distance between the two points is $r - 10$ and the maximum distance is $r + 10$ .

Given $min ∣ z_{1} - z_{2} ∣ = r - 10 = 2$ , so $r = 12$ .

$∴ max ∣ z_{1} - z_{2} ∣ = r + 10 = 22$ .

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →