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(A) The curves are $x^2+y^2=25$ (a circle with radius $5$) and $y=|x-1|$.Intersection points:For $y=x-1$,$x^2+(x-1)^2=25 \Rightarrow 2x^2-2x-24=0 \Rig

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$77$

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(A) The curves are $x^2+y^2=25$ (a circle with radius $5$) and $y=|x-1|$.Intersection points:For $y=x-1$,$x^2+(x-1)^2=25 \Rightarrow 2x^2-2x-24=0 \Rightarrow x^2-x-12=0 \Rightarrow (x-4)(x+3)=0$. Since $y \ge 0$,we take $x=4$,$y=3$.For $y=-(x-1)$,$x^2+(-x+1)^2=25 \Rightarrow 2x^2-2x-24=0 \Rightarrow x=-3$,$y=4$.The area of the smaller region bounded by the circle and the lines is the area of the circular sector plus the triangle formed by the lines.The area of the smaller region $A_s = \int_{-3}^4 (\sqrt{25-x^2} - |x-1|) dx$.Alternatively,using the geometry of the circle,the area of the smaller region is the sum of the area of the triangle with vertices $(1,0), (4,3), (-3,4)$ and the circular sectors.The area of the larger region $A_L = 	ext{Total Area} - A_s = 25\pi - A_s$.Calculating the area of the smaller region $A_s = \int_{-3}^4 \sqrt{25-x^2} dx - \int_{-3}^4 |x-1| dx$.$int_{-3}^4 \sqrt{25-x^2} dx = [\frac{x}{2}\sqrt{25-x^2} + \frac{25}{2}\sin^{-1}(\frac{x}{5})]_{-3}^4 = (2(3) + \frac{25}{2}\sin^{-1}(\frac{4}{5})) - (-\frac{3}{2}(4) + \frac{25}{2}\sin^{-1}(-\frac{3}{5})) = 6 + 6 + \frac{25}{2}(\sin^{-1}(\frac{4}{5}) + \sin^{-1}(\frac{3}{5})) = 12 + \frac{25}{2}(\frac{\pi}{2}) = 12 + \frac{25\pi}{4}$.$int_{-3}^4 |x-1| dx = \int_{-3}^1 (1-x) dx + \int_{1}^4 (x-1) dx = [x-\frac{x^2}{2}]_{-3}^1 + [\frac{x^2}{2}-x]_1^4 = (1-\frac{1}{2} - (-3-\frac{9}{2})) + (8-4 - (\frac{1}{2}-1)) = (0.5 + 7.5) + (4 + 0.5) = 8 + 4.5 = 12.5 = \frac{25}{2}$.$A_s = 12 + \frac{25\pi}{4} - \frac{25}{2} = \frac{25\pi}{4} - 0.5$.$A_L = 25\pi - (\frac{25\pi}{4} - 0.5) = \frac{75\pi}{4} + 0.5 = \frac{75\pi+2}{4} = \frac{1}{4}(75\pi+2)$.Thus,$b=75, c=2$.$b+c = 75+2 = 77$.

If the area of the larger portion bounded between the curves $x^2+y^2=25$ and $y=|x-1|$ is $\frac{1}{4}(b \pi+c)$,where $b, c \in N$,then $b+c$ is equal to $ . . .. .. $

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