What is the area of the circle defined by x^2 + 8x + y^2 - 14y + 61 = 0 that lies beneath the line y=7 ?
Complete the square on x , y
x^2 +8x + 16 + y^2 - 14y + 49 = 16 + 49
( x + 4)^2 + ( y - 7)^2 = 65
The center is ( - 4, 7)
The line y = 7 will create two half- circles
So (1/2) of the area of the circle will lie below y =7
