Let a circle C touch the lines L1:4x−3y+K1=0 and L2:4x−3y+K2=0,K1,K2∈R. If a line passing through the centre of the circle C intersects L1 at (−1,2) and L2 at (3,−6), then the equation of the circle C is
(A)(x−1)2+(y−2)2=4
(B)(x+1)2+(y−2)2=4
(C)(x−1)2+(y+2)2=16✓ Correct
(D)(x−1)2+(y−2)2=16
Answer: (C)
Solution
The circle touches two parallel lines 4x−3y+K1=0 and 4x−3y+K2=0. Since (−1,2) lies on L1: −4−6+K1=0⇒K1=10. Since (3,−6) lies on L2: 12+18+K2=0⇒K2=−30. The distance between the parallel lines is the diameter: 2r=5∣10−(−30)∣=8, so r=4. The center is the midpoint of (−1,2) and (3,−6): (2−1+3,22−6)=(1,−2). Thus the equation is (x−1)2+(y+2)2=16.
JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option