The greatest distance of the point  P (10, 7)  from the circle x2+y2−4x−2y−20=0, is

We have,S1=102+72−4×10−2×7−20>0So, P lies outside the circle. Join P with the centre C (2, 1) of the given circle. Suppose PC cuts the circle at A and B. Then, PB is the greatest distance of P from the circle.Now,  we have PC=(10−2)2+(7−1)2=10and,  CB= Radius=4+1+20=5∴ PB=PC+CB=(10+5)=15