JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The first equation |z-(4+8i)|= 10 represents a circle with center C(4, 8) and radius r = 10 . The second equation |z-(3+5i)|+|z-(5+11i)|=4 5 represents an ellipse with foci at S_1(3, 5) and S_2(5, 11), and length of the major axis 2a = 4 5 . The distance between the foci is 2ae = (5-3)^2 + (11-5)^2 = 4+36 = 40 = 2 10 . The center of the ellipse is the midpoint of the line segment joining the foci, which is ( 3+5 2 , 5+11 2 ) = (4, 8). This is the same as the center of the circle. The semi-major axis is a = 2 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 = 10 . Since the circle and the ellipse are concentric and the radius of the circle r = 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