JEE Main 2026 — Mathematics Algebra

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

The second equation $|z-(3+5i)|+|z-(5+11i)|=4\sqrt{5}$∣z−(3+5i)∣+∣z−(5+11i)∣=45​ represents an ellipse with foci at $S_1(3, 5)$S1​(3,5) and $S_2(5, 11)$S2​(5,11), and length of the major axis $2a = 4\sqrt{5}$2a=45​.

The distance between the foci is $2ae = \sqrt{(5-3)^2 + (11-5)^2} = \sqrt{4+36} = \sqrt{40} = 2\sqrt{10}$2ae=(5−3)2+(11−5)2​=4+36​=40​=210​.

The center of the ellipse is the midpoint of the line segment joining the foci, which is $\left(\dfrac{3+5}{2}, \dfrac{5+11}{2}\right) = (4, 8)$(23+5​,25+11​)=(4,8). This is the same as the center of the circle.

The semi-major axis is $a = 2\sqrt{5}$a=25​, so $a^2 = 20$a2=20.

The semi-minor axis $b$b is given by $b^2 = a^2 - (ae)^2 = 20 - 10 = 10$b2=a2−(ae)2=20−10=10, which gives $b = \sqrt{10}$b=10​.

Since the circle and the ellipse are concentric and the radius of the circle $r = \sqrt{10}$r=10​ is equal to the semi-minor axis $b$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$2 points of intersection, meaning there are $2$2 values of $z$z satisfying both equations.

Answer: $2$2