Let the tangents drawn from the origin to the circle, x2+y2−8x−4y+16=0 touch it at the points A and B. Then (AB)2 is equal to:
(A)556
(B)532
(C)564✓ Correct
(D)552
Answer: (C)
Solution
The circle x2+y2−8x−4y+16=0 has centre (4,2) and radius 2. The origin lies at distance OS=16+4=25 from the centre, so the tangent length is OT=OS2−R2=20−4=4. The chord of contact AB has length OS2R⋅OT=252(2)(4)=58, so (AB)2=564.
JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option