JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The characteristic equation is given by (A- I) = 0. bmatrix 1- & 2 & 7 \\ 4 & -2- & 8 \\ 3 & 8 & -7- bmatrix = 0 Expanding the determinant along the first row: (1- )[(-2- )(-7- ) - 64] - 2[4(-7- ) - 24] + 7[32 - 3(-2- )] = 0 (1- )( ^2 + 9 - 50) - 2(-4 - 52) + 7(3 + 38) = 0 - ^3 - 8 ^2 + 59 - 50 + 8 + 104 + 21 + 266 = 0 ^3 + 8 ^2 - 88 - 320 = 0 By inspection, = 8 is a root since 8^3 + 8(8^2) - 88(8) - 320 = 512 + 512 - 704 - 320 = 0. Factoring out ( - 8), we get: ( - 8)( ^2 + 16 + 40) = 0 The roots are = 8 and = -16 256 - 160 2 = -8 2 6 . The largest possible value of is p = 8. Substituting p = 8 into the given equation of the circle: (x-8)^2 + (y-16)^2 = 320 To find the intersection points with the x-axis, substitute y = 0: (x-8)^2 + (-16)^2 = 320 (x-8)^2 + 256 = 320 (x-8)^2 = 64 x-8 = 8 x = 0 or x = 16 The circle intersects the x-axis at (0, 0) and (16, 0). To find the intersection points with the y-axis, substitute x = 0: (-8)^2 + (y-16)^2 = 320 64 + (y-16)^2 = 320 (y-16)^2 = 256 y-16 = 16 y = 0 or y = 32 The circle intersects the y-axis at (0, 0) and (0, 32). The distinct points of intersection with the coordinate axes are (0, 0), (16, 0), and (0, 32). Thus, the circle intersects the coordinate axes at exactly 3 points. Answer: 3 points