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?

mathleteig Dec 20, 2019

#1**+1 **

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.

Guest Dec 20, 2019

#2**+2 **

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

CPhill Dec 20, 2019