+1911 Help I'm really stuck on this
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 x^2-14x+3y+70=15+9y-y^2 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
This is a circle with a center of ( 7, 3) and a radius of sqrt 3
The total area is formed by a triangle and an arc in the circle of 360° - 131.81° = 228.19°
The area of the triangle = (1/2) (sqrt 3)^2 * sin (131.81°) ≈ 1.118
The area of the arc is pi (sqrt 3)^2 (228.19 / 360) ≈ 5.974
The total area ≈ 1.118 + 5.974 ≈ 7.092
