+884 What is the area of the circle defined by \(x^2-6x +y^2-14y +33=0\) that lies beneath the line y=7 ?
+128408 Rearrange as
x^2 -6y +y^2 - 14y = -33 complete the square on x and y
x^2 - 6y + 9 + y^2 - 14y + 49 = -33 + 9 + 49
(x - 3)^2 + (y - 7)^2 = 25
This is a circle centered at ( 3, 7) with a radius of 5
The part lying below y = 7 is the area of 1/2 of this circle =
pi * 5^2 / 2 =
(25/2)pi ≈ 39.3 units^2
