Let a circle pass through the origin and its centre be the point of intersection of two mutually perpendicular lines x + (k-1)y + 3 = 0 and 2x + k^2 y - 4 = 0 . If the line x - y + 2 = 0 intersects the circle at the points A and B, then (AB)^2 is equal to:

Step-by-step solution

The given lines are x + (k-1)y + 3 = 0 and 2x + k^2 y - 4 = 0 . Since they are mutually perpendicular, the product of their slopes is -1 : ( -1 k-1 ) ( -2 k^2 ) = -1 2 k^2(k-1) = -1 k^3 - k^2 + 2 = 0 By trial, k = -1 is a root. Factoring gives (k+1)(k^2 - 2k + 2) = 0 . Since k^2 - 2k + 2 = 0 has no real roots, k = -1 . Substituting k = -1 into the equations of the lines, we get: x - 2y + 3 = 0 2x + y - 4 = 0 Solving these two equations, we get the point of intersection as x = 1 and y = 2 . Thus, the centre of the c