JEE Main 2026 Apr 8 Shift 2, Mathematics Q2: Rotation And Geometric Applications

The second equation describes an ellipse with foci $S_1(3,5)$ and $S_2(5,11)$, since $|z-S_1|+|z-S_2|$ is constant.

$2ae=|S_1{S}_2|=\sqrt{4+36}=2\sqrt{10}$ and $2a=4\sqrt{5}$, giving $a=2\sqrt{5}$ and $e=\frac{1}{\sqrt{2}}$.

Then $b^2=a^2(1-e^2)=20\cdot \frac{1}{2}=10$, so the semi-minor axis is $\sqrt{10}$.

The centre of the ellipse is the midpoint of the foci, which is $(4,8)$.

The first equation $|z-(4+8i)|=\sqrt{10}$ is a circle centred at $(4,8)$ with radius $\sqrt{10}$.

So the circle and ellipse are concentric, and the circle's radius equals the ellipse's semi-minor axis. Geometrically the circle touches the ellipse at the two endpoints of the minor axis.

$\therefore$ There are exactly $2$ common points.