The least distance of the point (10,7) from the circle x^2+y^2-4 x-2 y-20=0 is

Step-by-step solution

Given equation (x^2+y^2-4 x-2 y-20=0 ) So, C=(2,1) and radius = (g)^2+(f)^2-c = (2)^2+(1)^2+20 Radius =5 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 (10-2)^2+(7-1)^2 =10 units So, the minimum distance from the circle therefore becomes 10-5=5 units