JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let the foot of the perpendicular from the origin to the chord AB be P(h, k). The slope of OP is k h . Since the chord AB is perpendicular to OP, its equation is given by: y - k = - h k (x - h) hx + ky = h^2 + k^2 This can be rewritten as hx + ky h^2 + k^2 = 1. The chord AB subtends a right angle at the origin. We homogenize the equation of the circle x^2 + y^2 - 6x - 8y - 11 = 0 with the equation of the chord to find the joint equation of the lines OA and OB: x^2 + y^2 - (6x + 8y) ( hx + ky h^2 + k^2 ) - 11 ( hx + ky h^2 + k^2 )^2 = 0 Since the lines OA and OB are perpendicular, the sum of the coefficients of x^2 and y^2 in this homogenized equation must be zero. Coefficient of x^2: 1 - 6h h^2 + k^2 - 11h^2 (h^2 + k^2)^2 Coefficient of y^2: 1 - 8k h^2 + k^2 - 11k^2 (h^2 + k^2)^2 Equating the sum of these coefficients to zero: 2 - 6h + 8k h^2 + k^2 - 11(h^2 + k^2) (h^2 + k^2)^2 = 0 2 - 6h + 8k h^2 + k^2 - 11 h^2 + k^2 = 0 2(h^2 + k^2) - 6h - 8k - 11 = 0 h^2 + k^2 - 3h - 4k - 11 2 = 0 Replacing (h, k) with (x, y), the locus of the foot of the perpendicular is: x^2 + y^2 - 3x - 4y - 11 2 = 0 Comparing this with the given locus equation x^2 + y^2 - x - y - = 0, we get: = 3, = 4, = 11 2 Therefore, the value of + + 2 is: 3 + 4 + 2 ( 11 2 ) = 7 + 11 = 18 Answer: 18