If the area of the larger portion bounded between the curves $x^2+y^2 = 25$ and $y = |x-1|$ is $\dfrac{1}{4}(b\pi+c)$, $b,c\in\mathbb{N}$, then $b+c$ is equal to
Step-by-Step Solution
Key Concept: Find where $y=|x-1|$ intersects $x^2+y^2=25$ (at $x=-3,y=4$ and $x=4,y=3$), compute the circle sector area and subtract/add the triangular portions to get the larger region.
Intersections of $y=x-1$ with circle: $x^2+(x-1)^2=25 \Rightarrow x=4$; of $y=-x+1$: $x=-3$.
Area of larger portion $= $ (Circle area) $-$ (smaller region).
Smaller region $= \int_{-3}^{4}\sqrt{25-x^2}\,dx - \tfrac{1}{2}\times 4\times 4 - \tfrac{1}{2}\times 3\times 3$.
Computing: $\int_{-3}^{4}\sqrt{25-x^2}dx = \tfrac{75\pi}{4}+\tfrac{1}{2}$.
Larger area $= 25\pi - \left(\tfrac{75\pi}{4}+\tfrac{1}{2}-8-\tfrac{9}{2}\right) = \tfrac{75\pi+2}{4} = \tfrac{1}{4}(75\pi+2)$.
So $b=75$, $c=2$, $b+c = 77$.
Correct Answer: 77