COMEDK Exam 2016 Solved Paper

(*) Let point P(x1,y1) be any point on the circle, therefore it satisfy the circle
(x1−3)2+(y1+2)2=5r2 . . . (i)
The length of the tangent drawn from point p(x1,y1) to the circle (x−3)2+(y+2)2=r2 is
‌√(x1−3)2+(y1+2)2−r2=√5r2−r2
‌=√4r2=2r
⇒‌2r=16
⇒‌r=8
So, area between two circles
‌=π(5r2)−πr2
‌=5πr2−πr2
‌=4πr2=4π×64
‌=256π‌ sq units ‌