A circle with centre C has equation: x^2 + y^2 + 20x - 14 y + 49 = 0. Express the circle in the form (x-a)^2 +(y-b)^2=r^2. Show that the circle touches the y-axis and crosses the x-axis in two distinct points.
Firstly we rearrange the expression to the required form:x2 + y2 + 20x - 14y + 49 = 0 which gives (x + 10)2 - 100 + (y - 7)2 - 49 + 49 = 0 and so (x - (-10))2 + (y -7)2 = 102Hence a = -10, b = 7 and r = 10.Secondly to show that the circle touches the y-axis we consider when x = 0.(x + 10)2 + (y - 7)2 = 100, and once we substitute x = 0 we get: 102 + (y - 7)2 = 100(y - 7)2 = 0. Hence y = 7 (twice) and so the y axis is tangential to the circle.Finally to show that the circle crosses the x-axis at 2 distinct points we consider when y = 0.(x + 10)2 + (y - 7)2 = 100, and once we substitute y = 0 we get: (x + 10)2 + 49 = 100 which gives (x + 10)2 = 51.Which gives that:x + 10 = sqrt(51) which gives x = -10 + sqrt(51)and x + 10 = -sqrt(51) which gives x = -10 - sqrt(51)These roots are real and distinct and hence the circle crosses the x-axis at two distinct points as required.
