JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let the coordinates of point B be (h, k). The center of the circle with diameter AB is C_1 ( h+3 2 , k 2 ) and its radius is r_1 = 1 2 (h-3)^2 + k^2 . The given circle is x^2 + y^2 = 36, which has center C_2(0, 0) and radius r_2 = 6. Since the circles touch internally, the distance between their centers is equal to the difference of their radii: C_1C_2 = r_2 - r_1 ( h+3 2 )^2 + ( k 2 )^2 = 6 - 1 2 (h-3)^2 + k^2 Multiplying the entire equation by 2, we get: (h+3)^2 + k^2 + (h-3)^2 + k^2 = 12 This equation represents the locus of a point B(h, k) such that the sum of its distances from two fixed points S_1(-3, 0) and S_2(3, 0) is constant and equal to 12. This is the standard definition of an ellipse. Thus, the foci of the ellipse are ( 3, 0) and the length of the major axis is 2a = 12. The distance between the foci is 2ae = 6. Substituting 2a = 12, we get 12e = 6 e = 1 2 . We need to find the value of 72e^2: 72e^2 = 72 ( 1 2 )^2 = 72 1 4 = 18 Answer: 18