JEE Main 2025 (Online) 23rd January Morning ShiftQuestion 1
If the area of the larger portion bounded between the curves $x^{2}+y^{2}=25$ and $y=|x-1|$ is $\dfrac{1}{4}(b\pi+c)$, $b,c\in\mathbf{N}$, then $b+c$ is equal to $\_\_\_\_\_\_\_\_$ .
Application of Integrals
Answer$77$
Solution
$y=|x-1|$ meets the circle at $(-3,4)$ and $(4,3)$. The part above it has area $\int_{-3}^4\left(\sqrt{25-x^2}-|x-1|\right)dx=\frac{25\pi}4-\frac12$, so the larger part is $25\pi-\frac{25\pi}4+\frac12=\frac14(75\pi+2)$ and $b+c=77$.