JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given circle C: (x-4)^2 + (y+3)^2 = 9 has its center at C_0(4, -3) and radius r = 3. The line QR has the equation x - y = 4, which has a slope of 1. Since PQ = PR, the point P( , ) must lie on the perpendicular bisector of the chord QR. The perpendicular bisector of any chord of a circle passes through its center. Therefore, the perpendicular bisector passes through C_0(4, -3) and has a slope of -1 (since it is perpendicular to QR). The equation of the perpendicular bisector is: y - (-3) = -1(x - 4) x + y = 1 Since P( , ) lies on this perpendicular bisector, we have: + = 1 = 1 - Also, P( , ) lies on the circle C, so its coordinates must satisfy the circle's equation: ( - 4)^2 + ( + 3)^2 = 9 Substituting = 1 - into the equation: ( - 4)^2 + (1 - + 3)^2 = 9 ( - 4)^2 + (4 - )^2 = 9 2( - 4)^2 = 9 - 4 = 3 2 = 4 3 2 We need to find the value of (6 + 8 )^2. Substituting = 1 - : 6 + 8 = 6 + 8(1 - ) = 8 - 2 Now, substituting the value of : 8 - 2 = 8 - 2 (4 3 2 ) = 6 2 = 3 2 Squaring this value gives: (6 + 8 )^2 = ( 3 2 )^2 = 18 Answer: 18