Let $A=\{z \in \mathbb{C}:|z-2| \leq 4\}$ and $\mathrm{B}=\{\mathrm{z} \in \mathbb{C}:|\mathrm{z}-2|+|\mathrm{z}+2|=5\}$. Then the max $\left\{\left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|: \mathrm{z}_{1} \in \mathrm{~A}\right.$ and $\left.\mathrm{z}_{2} \in \mathrm{~B}\right\}$ is

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Complex Numbers

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}

Answer: (C)

The set

A

is the closed disc centred at

(2, 0)

with radius

4

. The set

B

is the ellipse with foci

(\pm 2, 0)

and

∣ z - 2∣ + ∣ z + 2∣ = 5

, so

2 a = 5

, its centre is the origin, and its vertices on the axis are

(\pm \frac{5}{2}, 0)

. For any ellipse point

q

, the farthest disc point lies at distance

∣ q - (2, 0) ∣ + 4

, and

∣ q - (2, 0) ∣

is greatest at the left vertex

q = (- \frac{5}{2}, 0)

, where it equals

\frac{9}{2}

. Taking

z_{2} = (- \frac{5}{2}, 0)

and

z_{1} = (6, 0)

max ∣ z_{1} - z_{2} ∣ = 6 - (- \frac{5}{2}) = \frac{17}{2} .

The maximum value is

\frac{17}{2}

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option