Curves involved

We have:

• Circle: $x^2+y^2=25$ (center $(0,0)$, radius $5$)
• Curve: $y=|x-1|$, i.e. $y=x-1 \quad (x\ge 1), \qquad y=1-x \quad (x\le 1)$

This is a V-shaped graph with vertex at $(1,0)$.

Find intersection points with the circle

We solve separately.

(i) With $y=x-1$: $x^2+(x-1)^2=25$ $2x^2-2x+1=25$ $2x^2-2x-24=0$ $x^2-x-12=0$ $(x-4)(x+3)=0$ So $x=4$ or $x=-3$. But for branch $y=x-1$, we need $x\ge 1$, so only $x=4$ works. Then $y=3$.

(ii) With $y=1-x$: $x^2+(1-x)^2=25$ $2x^2-2x-24=0$ $x=4 \text{ or } -3$ For branch $y=1-x$, we need $x\le 1$, so only $x=-3$ works. Then $y=4$.

Hence the intersection points are: $(4,3),\quad (-3,4)$

What region is asked?

The line pair $y=|x-1|$ cuts the circle into two portions:

• a smaller region near the top, bounded by the arc from $(-3,4)$ to $(4,3)$ and the two line segments joining these points to $(1,0)$,
• the larger remaining part of the disk.

So, $\text{Area of larger portion} = \text{Area of circle} - \text{Area of smaller portion}.$

Area of the smaller portion

The smaller portion is inside the circle and above $y=|x-1|$.

So its area is $A_s=\int_{-3}^{1}\left(\sqrt{25-x^2}-(1-x)\right)dx + \int_{1}^{4}\left(\sqrt{25-x^2}-(x-1)\right)dx.$

Combine the linear parts: $A_s=\int_{-3}^{4}\sqrt{25-x^2}\,dx - \left[\int_{-3}^{1}(1-x)dx+\int_{1}^{4}(x-1)dx\right].$

Compute the straight-line part

$\int_{-3}^{1}(1-x)dx=\left[x-\frac{x^2}{2}\right]_{-3}^{1}$ At $x=1$: $1-\frac12=\frac12$

At $x=-3$: $-3-\frac92=-\frac{15}{2}$

So, $\int_{-3}^{1}(1-x)dx=\frac12-\left(-\frac{15}{2}\right)=8.$

Next, $\int_{1}^{4}(x-1)dx=\left[\frac{x^2}{2}-x\right]_{1}^{4}$ At $x=4$: $8-4=4$

At $x=1$: $\frac12-1=-\frac12$

Hence, $\int_{1}^{4}(x-1)dx=4-\left(-\frac12\right)=\frac92.$

Therefore total line-area removed is $8+\frac92=\frac{25}{2}.$

Compute $\int_{-3}^{4}\sqrt{25-x^2}\,dx$

Use the standard formula: $\int \sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+C.$

Here $a=5$, so $I=\int_{-3}^{4}\sqrt{25-x^2}\,dx$ $=\left[\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2}\sin^{-1}\left(\frac{x}{5}\right)\right]_{-3}^{4}.$

At $x=4$: $\frac{4}{2}\cdot 3 + \frac{25}{2}\sin^{-1}\left(\frac45\right)=6+\frac{25}{2}\sin^{-1}\left(\frac45\right).$

At $x=-3$: $\frac{-3}{2}\cdot 4 + \frac{25}{2}\sin^{-1}\left(\frac{-3}{5}\right)=-6-\frac{25}{2}\sin^{-1}\left(\frac35\right).$

Therefore, $I=12+\frac{25}{2}\left(\sin^{-1}\frac45+\sin^{-1}\frac35\right).$

Now note that if $\theta=\sin^{-1}(4/5)$ and $\phi=\sin^{-1}(3/5)$, then $\theta+\phi=\frac{\pi}{2}$ because these are complementary acute angles of the $3$-$4$-$5$ triangle.

Hence, $I=12+\frac{25}{2}\cdot \frac{\pi}{2}=12+\frac{25\pi}{4}.$

Area of smaller portion

$A_s=\left(12+\frac{25\pi}{4}\right)-\frac{25}{2}$ $=\frac{25\pi}{4}-\frac12.$

Area of larger portion

Total area of circle: $25\pi.$

So, $A_L=25\pi-A_s$ $=25\pi-\left(\frac{25\pi}{4}-\frac12\right)$ $=\frac{75\pi}{4}+\frac12$ $=\frac{1}{4}(75\pi+2).$

Therefore, $b=75,\quad c=2$ and $b+c=77.$

Comparison with stored answer

Stored correct answer = $77$.

Our derived answer also equals $77$, so it agrees.