JEE Main 2026 — Jan 28, Shift 2 — Mathematics solution

JEE Main2026Jan 28, Shift 2Mathematics

Let $A = {z \in C : ∣ z - 2∣ \leq 4}$ and $B = {z \in C : ∣ z - 2∣ + ∣ z + 2∣ = 5}$ . Then the max ${∣ z_{1} - z_{2} ∣ : z_{1} \in A and z_{2} \in B}$ is

A
$\frac{15}{2}$
B
8
C
$\frac{17}{2}$
D
9

Solution

$A$ is the closed disc centred at $(2, 0)$ with radius $4$ .

$B$ is the ellipse with foci $(2, 0)$ and $(- 2, 0)$ and sum of focal distances $= 5$ , so $2 a = 5 \Rightarrow a = \frac{5}{2}$ . Centre $(0, 0)$ . Vertices on the major axis at $(\pm \frac{5}{2}, 0)$ .

Maximum $∣ z_{1} - z_{2} ∣$ is the largest distance between any point of the disc and any point of the ellipse. The disc reaches farthest to the right at $(6, 0)$ ; the ellipse reaches farthest to the left at $(- \frac{5}{2}, 0)$ .

Max distance $= 6 - (- \frac{5}{2}) = \frac{17}{2}$ .

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