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:
C. 18
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 circle is C(1, 2). Since the circle passes through the origin (0,0), the square of its radius R is: R^2 = (1 - 0)^2 + (2 - 0)^2 = 5 The perpendicular distance d from the centre C(1, 2) to the line x - y + 2 = 0 is: d = |1 - 2 + 2| 1^2 + (-1)^2 = 1 2 The length of the chord AB is given by 2 R^2 - d^2 . Therefore, its square is: (AB)^2 = 4(R^2 - d^2) = 4 (5 - 1 2 ) = 4 9 2 = 18 Answer: 18
Related: Mathematics — Circle · All PYQ Banks