JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Complex Numbers

The second condition describes an ellipse with foci $S_1(3,5)$ and $S_2(5,11)$. The focal separation is $\sqrt{(5-3{)}^2+(11-5{)}^2}=2\sqrt{10}$, so the focal distance is $c=\sqrt{10}$, while $2a=4\sqrt{5}$ gives $a^2=20$; hence $b^2=a^2-c^2=10$. The centre is the midpoint $(4,8)$. The first condition is a circle centred at the same point $(4,8)$ with radius $\sqrt{10}$, which equals the semi-minor axis $b$. A point of the ellipse lies at distance $b$ from the centre only at the two ends of the minor axis, so the circle meets the ellipse in exactly $2$ points.