JEE Main 2026 — Complex Number Question with Solution

The first equation $|z-(4+8i)|=\sqrt{10}$ represents a circle with center $C(4,8)$ and radius $r=\sqrt{10}$.

The second equation $|z-(3+5i)|+|z-(5+11i)|=4\sqrt{5}$ represents an ellipse with foci at $S_1(3,5)$ and $S_2(5,11)$, and length of the major axis $2a=4\sqrt{5}$.

The distance between the foci is $2ae=\sqrt{(5-3{)}^2+(11-5{)}^2}=\sqrt{4+36}=\sqrt{40}=2\sqrt{10}$.

The center of the ellipse is the midpoint of the line segment joining the foci, which is $\left(\frac{3+5}{2},\frac{5+11}{2}\right)=(4,8)$. This is the same as the center of the circle.

The semi-major axis is $a=2\sqrt{5}$, so $a^2=20$.

The semi-minor axis $b$ is given by $b^2=a^2-(ae{)}^2=20-10=10$, which gives $b=\sqrt{10}$.

Since the circle and the ellipse are concentric and the radius of the circle $r=\sqrt{10}$ is equal to the semi-minor axis $b$ of the ellipse, the circle is inscribed in the ellipse and touches it exactly at the two endpoints of the minor axis.

Therefore, there are exactly $2$ points of intersection, meaning there are $2$ values of $z$ satisfying both equations.

Answer: $2$