If the area of the larger portion bounded between the curves $x^2+y^2=25$ and $y=|x-1|$ is $\frac14(b\pi+c)$, $b,c\in N$, then $b+c$ is equal to

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] The lines $y=x-1$ and $y=1-x$ meet the circle at $(4,3)$ and $(-3,4)$. The smaller portion (above the V) has area $\int_{-3}^4\left(\sqrt{25-x^2}-|x-1|\right)dx=\left(12+\frac{25\pi}4\right)-\frac{25}2=\frac{25\pi-2}4$. The larger portion is $25\pi-\frac{25\pi-2}4=\frac{75\pi+2}4$, so $b=75$, $c=2$ and $b+c=77$. Answer: 77.

Q. If the area of the larger portion bounded between the curves $x^2+y^2=25$ and $y=|x-1|$ is $\frac14(b\pi+c)$, $b,c\in N$, then $b+c$ is equal to

Explanation

Reviewed by Dr. Abubacker

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