JEE Main 2026 (Online) 28th January Evening ShiftQuestion 1
Let $A=\{z\in\mathbf{C}:|z-2|\le4\}$ and $B=\{z\in\mathbf{C}:|z-2|+|z+2|=5\}$. Then the $\max\{|z_1-z_2|:z_1\in A$ and $z_2\in B\}$ is :
Complex Numbers
Solution
$B$ is an ellipse with foci $\pm2$, $a=\frac52$. Max of $|z_2-2|$ over $B$ is $a+c=\frac92$ (at $z=-\frac52$). Max distance $=\frac92+4=\frac{17}{2}$.