+1439 Find the area of the region enclosed by the graph of the equation $x^2-14x+3y+70=15+9y-y^2$ that lies below the line $y=x-3$.
+129895 Simplify as
x^2 -14x + y^2 -6y = - 55 complete the square on x, y
x^2 -14x + 49 + y^2 -6y + 9 = -55 + 49 + 9
(x - 7)^2 + (y - 3)^2 = 3
y = sqrt ( 3 - (x-7)^2 ) + 3
This is a circle centered at (7,3) with a rdius of sqrt 3
This is probably best approximated with a graph
The area we want is the area of triangle ABC plus the area of major sector BC
(1/2) (sqrt 3)^2 sin (131.81°) + pi * (sqrt 3)^2 * ( 360 - 131.81) / 360 ≈ 7.09
