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+k2y−4=0. If the line x−y+2=0 intersects the circle at the points A and B, then (AB)2 is equal to :
(A)10
(B)27
(C)18✓ Correct
(D)34
Answer: (C)
Solution
The lines are perpendicular when the product of their slopes is −1: k−1−1⋅k2−2=−1, which simplifies to k3−k2+2=0 with real root k=−1. The lines become x−2y+3=0 and 2x+y−4=0, meeting at the centre (1,2). As the circle passes through the origin, r2=12+22=5. The distance from (1,2) to x−y+2=0 is 2∣1−2+2∣=21, so
AB=2r2−d2=25−21=32,
giving (AB)2=18.
JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option