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

We are asked to find the area of the larger portion bounded between the circle $$x^2 + y^2 = 25$$ and the curve $$y = |x - 1|$$, express it as $$\frac{1}{4}(b\pi + c)$$ with $$b,c\in\mathbb{N}$$, and then determine $$b+c$$.

The circle centered at $$(0,0)$$ has radius 5, and the curve $$y = |x - 1|$$ is a V-shaped graph with vertex at $$(1,0)$$, lying entirely in the upper half-plane. To find their intersection points, substitute $$y = |x-1|$$ into $$x^2 + y^2 = 25$$, which gives

$$x^2 + (x-1)^2 = 25 \implies 2x^2 - 2x + 1 = 25 \implies 2x^2 - 2x - 24 = 0 \implies x^2 - x - 12 = 0 \implies (x - 4)(x + 3) = 0,$$

so $$x = 4$$ or $$x = -3$$. The intersection points are therefore $$(4,3)$$ and $$(-3,4)$$.

The smaller region is bounded above by the upper semicircle and below by the V-curve from $$x = -3$$ to $$x = 4$$. Its area is

$$A_{\text{small}} = \int_{-3}^{4}\sqrt{25 - x^2}\,dx \;-\;\int_{-3}^{4}|x-1|\,dx.$$

First, using the standard antiderivative $$\int \sqrt{a^2 - x^2}\,dx = \frac{x\sqrt{a^2 - x^2}}{2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$$ with $$a=5$$, we evaluate at the bounds:

At $$x=4$$: $$\frac{4\cdot3}{2} + \frac{25}{2}\sin^{-1}\!\frac{4}{5} = 6 + \frac{25}{2}\sin^{-1}\!\frac{4}{5},$$

At $$x=-3$$: $$\frac{-3\cdot4}{2} + \frac{25}{2}\sin^{-1}\!\frac{-3}{5} = -6 - \frac{25}{2}\sin^{-1}\!\frac{3}{5}.$$

Since $$\sin^{-1}\!\frac{4}{5} + \sin^{-1}\!\frac{3}{5} = \frac{\pi}{2}$$, it follows that

$$\int_{-3}^{4}\sqrt{25 - x^2}\,dx = \bigl(6 + \tfrac{25}{2}\sin^{-1}(4/5)\bigr) - \bigl(-6 - \tfrac{25}{2}\sin^{-1}(3/5)\bigr) = 12 + \frac{25\pi}{4}.$$

Next, split the integral of the V-curve at $$x=1$$:

$$\int_{-3}^{1}(1-x)\,dx = \Bigl[x - \tfrac{x^2}{2}\Bigr]_{-3}^{1} = \frac12 + \frac{15}{2} = 8,$$

$$\int_{1}^{4}(x-1)\,dx = \Bigl[\tfrac{x^2}{2} - x\Bigr]_{1}^{4} = (8 - 4) - \bigl(\tfrac12 - 1\bigr) = 4 + \tfrac12 = \frac{9}{2},$$

so that $$\int_{-3}^{4}|x-1|\,dx = 8 + \frac{9}{2} = \frac{25}{2}.$$

Therefore,

$$A_{\text{small}} = \Bigl(12 + \frac{25\pi}{4}\Bigr) - \frac{25}{2} = \frac{25\pi - 2}{4}.$$

Since the total area of the circle is $$25\pi$$, the larger region has area

$$A_{\text{larger}} = 25\pi - \frac{25\pi - 2}{4} = \frac{100\pi - 25\pi + 2}{4} = \frac{75\pi + 2}{4} = \frac{1}{4}(75\pi + 2).$$

Thus, $$b = 75$$ and $$c = 2$$, giving $$b + c = 77$$.