JEE Mathematics 2026 Question with Solution

Given: $A=\{z\in\mathbb{C}:|z-2|\le 4\}$ and $B=\{z\in\mathbb{C}:|z-2|+|z+2|=5\}$.

Find: The maximum value of $|z_1-z_2|$ for $z_1\in A$ and $z_2\in B$.

Concept: $|z-a|\le r$ represents a closed disc with centre $a$ and radius $r$, while $|z-a|+|z-b|=2c$ represents an ellipse with foci $a$ and $b$ and major axis length $2c$.

For $A$,

$$|z-2|\le 4$$

so it is a disc with centre $(2,0)$ and radius $4$. Hence its extreme points on the real axis are $6$ and $-2$.

For $B$,

$$|z-2|+|z+2|=5$$

so it is an ellipse with foci at $(2,0)$ and $(-2,0)$. Since

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

the vertices on the real axis are at $\pm \frac{5}{2}$.

Thus the farthest possible separation is obtained by taking the rightmost point of $A$ and the leftmost point of $B$:

$$|z_1-z_2|=6-\left(-\frac{5}{2}\right)=\frac{17}{2}$$

So the geometric maximum is $\frac{17}{2}$.

However, the solution concludes with Final Answer: $9$ and marks Option C as correct, which is inconsistent with the working shown above. Following the solution, the correct option is C.

Given: $A$ is a closed disc centred at $(2,0)$ with radius $4$, and $B$ is an ellipse with foci $(-2,0)$ and $(2,0)$ and major axis length $5$.

Find: The maximum distance between one point from $A$ and one point from $B$.

The set $A$ extends from $x=-2$ to $x=6$ on the real axis.

The set $B$ has semi-major axis

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

so its leftmost and rightmost vertices are

$$-\frac{5}{2} \quad \text{and} \quad \frac{5}{2}.$$

To maximize the distance between two points from these regions, choose boundary points in opposite directions along the real axis. Therefore take

$$z_1=6, \qquad z_2=-\frac{5}{2}.$$

Then

$$|z_1-z_2|=\left|6-\left(-\frac{5}{2}\right)\right|=\frac{17}{2}.$$

This shows the working in the solution's actually supports $\frac{17}{2}$, even though the page labels Option C and writes $9$ at the end. Because the source solution explicitly declares C, the extracted answer is recorded as C.