+4622 What is the area of the circle defined by \(x^2-6x +y^2-14y +33=0\) that lies beneath the line \( y=7\) ?
+129847 x^2 - 6x + y^2 - 14y + 33 = 0
x^2 - 6x + y^2 - 14y = -33 complete the square on x and y
x^2 - 6x + 9 + y^2 - 14y + 49 = -33 + 9 + 49 factor and simplify
(x - 3)^2 + (y - 7)^2 = 25
This a circle that is centered at (3, 7 ) with a radius of 5
So.....the area of the portion of the circle that lies below y = 7 is the area of the half circle
And this is = (1/2)* pi (5^2) = 12.5 pi units^2 ≈ 39.27 units^2
