Given equation (x2+y2−4x−2y−20=0) So, C=(2,1) and radius =√(g)2+(f)2−c ‌=√(2)2+(1)2+20 When substituted x=10 and y=7 in equation, then value becomes
(10)2+(7)2−4(10)−2(7)−20=75
which is greater than zero. Thus, the point (10,7) lies outside the circle. Its distance from the centre of the circle (2,1) is √(10−42+(7−1)2.=10 units So, the minimum distance from the circle therefore becomes 10−5=5 units