JEE Main 2026 — 28 January, Evening Shift — Question 18

$$A=\{z\in\mathbb C:\ |z-2|\le 4\}$$

is a closed disc with center $C_1=2$ and radius $4$.

$$B=\{z\in\mathbb C:\ |z-2|+|z+2|=5\}$$

is an ellipse with foci at $2$ and $-2$, and major axis length $5$.

Hence,

$$2a=5 \;\Rightarrow\; a=\frac{5}{2}$$

Distance between foci $=4$, so

$$c=2,\qquad b^2=a^2-c^2=\frac{25}{4}-4=\frac{9}{4}$$

Thus the farthest point of $B$ from the center $2$ lies at distance

$$\max |z-2| = a+c=\frac{5}{2}+2=\frac{9}{2}$$

To maximize $|z_1-z_2|$, take $z_1$ on $A$ farthest from $2$ in the opposite direction:

$$\max |z_1-z_2| = 4+\frac{9}{2}=\frac{17}{2}$$ $$\boxed{\frac{17}{2}}$$