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 $b, c \in \mathbb{N}$, then $b+c$ is equal to
Solution:
The intersection points of the circle $x^{2}+y^{2}=25$ and lines $y=|x-1|$ are found by solving $x^{2}+(x-1)^{2}=25$.
$$2x^{2}-2x-24=0 \Rightarrow x^{2}-x-12=0 \Rightarrow (x-4)(x+3)=0$$
So, the intersection points are at $x=4$ and $x=-3$.
The area of the larger portion is given by Total Area of Circle – Area of the smaller region between the curves.
$$A = 25\pi – \int_{-3}^{4} \left( \sqrt{25-x^{2}} – |x-1| \right) dx$$
$$A = 25\pi – \int_{-3}^{4} \sqrt{25-x^{2}} dx + \int_{-3}^{4} |x-1| dx$$
Calculating the integral of the lines (Area of triangles formed by $y=|x-1|$ on x-axis):
$$\int_{-3}^{4} |x-1| dx = \text{Area}(\Delta_1) + \text{Area}(\Delta_2) = \frac{1}{2}(4)(4) + \frac{1}{2}(3)(3) = 8 + 4.5 = 12.5$$
Calculating the integral of the circle using formula $\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$:
$$\int_{-3}^{4} \sqrt{25-x^{2}} dx = \left[ \frac{x}{2}\sqrt{25-x^{2}} + \frac{25}{2}\sin^{-1}\frac{x}{5} \right]_{-3}^{4}$$
$$= \left( 6 + \frac{25}{2}\sin^{-1}\frac{4}{5} \right) – \left( -6 + \frac{25}{2}\sin^{-1}\frac{-3}{5} \right)$$
$$= 12 + \frac{25}{2} \left( \sin^{-1}\frac{4}{5} + \sin^{-1}\frac{3}{5} \right) = 12 + \frac{25}{2}\left( \frac{\pi}{2} \right) = 12 + \frac{25\pi}{4}$$
Substituting these values back into the area equation:
$$A = 25\pi – \left( 12 + \frac{25\pi}{4} \right) + 12.5$$
$$A = 25\pi – \frac{25\pi}{4} + 0.5 = \frac{75\pi}{4} + \frac{2}{4} = \frac{1}{4}(75\pi + 2)$$
Comparing with $\frac{1}{4}(b\pi+c)$, we get $b=75$ and $c=2$.
$$b+c = 75 + 2 = 77$$
Ans. 77