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?
The circle has a radius of 12, and the line y = 6 - x splits it in half, so the area is 1/2*12^2*pi = 72*pi.
See the graph here : https://www.desmos.com/calculator/qeidqnvtts
The intersection of this line and the circle will form the diameter of the circle....to find the radius we can complete the square on x in the equation of the circle......so we have....
x^2 - 12x +36 + y^2 = 28 + 36
(x + 6)^2 + y^2 = 64
The center is (6, 0)
The radius = sqrt (64) = 8
So....the area we are looking for is
(1/2) pi (8)^2 =
32 pi units^2