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

Step 1: Identify the curves and intersection points

The first equation represents a circle centered at the origin with a radius of 5. The second equation represents a V-shaped graph with its vertex at $(1, 0)$, opening upwards. We need to find the intersection points of these two curves.

Step 2: Calculate intersection points

Substitute $y = |x-1|$ into the circle equation. This splits into two cases: $y = x-1$ for $x \ge 1$ and $y = -(x-1) = 1-x$ for $x < 1$. Solving $x^2 + (x-1)^2 = 25$ gives $2x^2 - 2x - 24 = 0$, or $x^2 - x - 12 = 0$, which factors to $(x-4)(x+3) = 0$. Since $x \ge 1$, we take $x=4$. Then $y = |4-1| = 3$. So, $(4, 3)$ is an intersection point. Solving $x^2 + (1-x)^2 = 25$ gives $2x^2 - 2x - 24 = 0$, or $x^2 - x - 12 = 0$. Since $x < 1$, we take $x=-3$. Then $y = |-3-1| = 4$. So, $(-3, 4)$ is another intersection point.

Step 3: Calculate the area of the smaller portion

The area of the smaller portion is the area under the V-shaped curve between $x=-3$ and $x=4$, minus the area of the triangle formed by the origin and the intersection points, plus the area of the sector. Alternatively, it's the area of the sector formed by the origin and the intersection points, minus the area of the triangle formed by the origin and the intersection points, plus the area under the line $y=|x-1|$. The area of the sector formed by the origin and the points $(-3{,}4)$ and $(4{,}3)$ can be found using the angle $\theta$ between the vectors $\vec{OA} = (-3{,}4)$ and $\vec{OB} = (4{,}3)$. $\cos \theta = \frac{(-3)(4) + (4)(3)}{5 \cdot 5} = \frac{-12+12}{25} = 0$, so $\theta = \frac{\pi}{2}$. The area of the sector is $\frac{1}{2}r^2\theta = \frac{1}{2}(5^2)(\frac{\pi}{2}) = \frac{25\pi}{4}$. The area of the triangle formed by the origin and the points $(-3{,}4)$ and $(4{,}3)$ is $\frac{1}{2} |x_1y_2 - x_2y_1| = \frac{1}{2} |(-3)(3) - (4)(4)| = \frac{1}{2} |-9 - 16| = \frac{25}{2}$. The area under $y=|x-1|$ from $x=-3$ to $x=4$ is $\int_{-3}^{1} (1-x) dx + \int_{1}^{4} (x-1) dx = [x - \frac{x^2}{2}]_{-3}^{1} + [\frac{x^2}{2} - x]_{1}^{4} = (1 - \frac{1}{2}) - (-3 - \frac{9}{2}) + (\frac{16}{2} - 4) - (\frac{1}{2} - 1) = \frac{1}{2} + \frac{15}{2} + 4 + \frac{1}{2} = \frac{17}{2} + 4 = \frac{25}{2}$. The area of the smaller region is the area of the sector minus the area of the triangle formed by the origin and the intersection points, plus the area under the V-shaped curve. This is $A_{\text{smaller}} = \frac{25\pi}{4} - \frac{25}{2} + \frac{25}{2} = \frac{25\pi}{4}$.

Step 4: Calculate the total area of the circle

The total area of the circle is given by the formula $\pi r^2$, where $r$ is the radius. With a radius of 5, the total area is $25\pi$.

Step 5: Calculate the area of the larger portion

The area of the larger portion is the total area of the circle minus the area of the smaller portion we just calculated. This gives us $25\pi - \frac{25\pi}{4} = \frac{100\pi - 25\pi}{4} = \frac{75\pi}{4}$.

Step 6: Compare with the given form and find b+c

We are given that the area of the larger portion is $\frac{1}{4}(b\pi + c)$. Comparing this with our calculated area $\frac{75\pi}{4}$, we can see that $b\pi + c = 75\pi$. Therefore, $b=75$ and $c=0$. However, the problem states $b, c \in \mathbb{N}$, which means $c$ must be a natural number (positive integer). This implies that the interpretation of the smaller area calculation might need adjustment. Let's re-evaluate the area bounded by the curves. The area bounded by the curves is the area of the circle minus the area of the region defined by $y=|x-1|$ inside the circle. The area of the region bounded by the curves is the area of the sector $AOB$ (where $A=(-3{,}4)$ and $B=(4{,}3)$) plus the area of the triangle $A(1{,}0)B$ plus the area of the triangle $A(-3{,}4)C$ where $C=(1{,}0)$ and $B(4{,}3)D$ where $D=(1{,}0)$. The area of the sector $AOB$ is $\frac{25\pi}{4}$. The area of the triangle $A(1{,}0)B$ is $\frac{1}{2} \text{base} \times \text{height}$. The base is the distance between $x=-3$ and $x=4$ along the line $y=0$, which is $4 - (-3) = 7$. The height is the $y$ -coordinate of the vertex $(1{,}0)$ relative to the line segment connecting $(-3{,}4)$ and $(4{,}3)$. This is not straightforward. Let's reconsider the area bounded by the curves. The area of the region bounded by the curves is the area of the circle minus the area of the region defined by $y=|x-1|$ inside the circle. The area of the region defined by $y=|x-1|$ inside the circle is the area of the triangle formed by the points $(-3{,}4)$, $(1{,}0)$, and $(4{,}3)$. This is a polygon. We can calculate its area using the shoelace formula or by dividing it into two triangles. Area of triangle with vertices $(-3{,}4)$, $(1{,}0)$, $(4{,}3)$: Using shoelace formula: $\frac{1}{2} |(-3)(0) + (1)(3) + (4)(4) - (4)(1) - (0)(4) - (3)(-3)|$ $= \frac{1}{2} |0 + 3 + 16 - 4 - 0 - (-9)|$ $= \frac{1}{2} |19 - 4 + 9| = \frac{1}{2} |24| = 12$. This is the area of the region *under* the V-shape. The area of the smaller portion is the area of the sector $AOB$ minus the area of the triangle $AOB$. Area of sector $AOB = \frac{25\pi}{4}$. Area of triangle $AOB = \frac{1}{2} |(-3)(3) - (4)(4)| = \frac{1}{2} |-9 - 16| = \frac{25}{2}$. So, the area of the smaller segment of the circle cut by the chord $AB$ is $\frac{25\pi}{4} - \frac{25}{2}$. The area of the region bounded by the curves is the area of the triangle formed by the points $(-3{,}4)$, $(1{,}0)$, and $(4{,}3)$ plus the area of the segment of the circle. Area bounded by curves = $12 + (\frac{25\pi}{4} - \frac{25}{2}) = \frac{25\pi}{4} + 12 - \frac{25}{2} = \frac{25\pi}{4} + \frac{24 - 25}{2} = \frac{25\pi}{4} - \frac{1}{2}$. This is the area of the region *inside* the circle and *above* the V-shape. This is the smaller portion. $A_{\text{smaller}} = \frac{25\pi}{4} - \frac{1}{2}$. $A_{\text{larger}} = A_{\text{circle}} - A_{\text{smaller}} = 25\pi - (\frac{25\pi}{4} - \frac{1}{2}) = 25\pi - \frac{25\pi}{4} + \frac{1}{2} = \frac{75\pi}{4} + \frac{1}{2}$. We are given $A_{\text{larger}} = \frac{1}{4}(b\pi + c)$. So, $\frac{75\pi}{4} + \frac{1}{2} = \frac{b\pi}{4} + \frac{c}{4}$. Comparing coefficients: $\frac{b\pi}{4} = \frac{75\pi}{4} \implies b = 75$. $\frac{c}{4} = \frac{1}{2} \implies c = 2$. Both $b=75$ and $c=2$ are natural numbers. Therefore, $b+c = 75+2 = 77$.

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