What is the area of the portion of the circle defined by \(x^2-12x+y^2=28\) that lies above the \(x\)-axis and to the right of the line \(y=6-x\)?
+128475 We could use either trig or calculus to solve this.....but a graphical solution seems WAY easier
Look at the graph here : https://www.desmos.com/calculator/cd8js1dkja
The line y = 6 - x forms a diameter with its intersection of the circle
Notice that the line y = x - 6 is perpendicular to y = 6- x....so.... it forms another diameter....and these lines divide the circle into 4 equal parts
So....the part lying to the right of y = 6 - x and above the x axis = (1/4) area of circle + (1/8) area of the circle =
(3/8) area of the circle
The radius = 8
So....the area we want is (3/8) * pi (8)^2 = 24 pi units^2
