JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Area Under Curves

The circle is $x^2+y^2=25$ and the curve $y=|x-1|$ has vertex $(1,0)$.

Intersection points: for $x\ge 1$, $y=x-1$: $x^2+(x-1{)}^2=25⟹2x^2-2x-24=0⟹x=4$ (the valid root in this branch).
For $x<1$, $y=1-x$: the same equation gives $x=-3$.

So the two curves meet at $(-3,4)$ and $(4,3)$. Since $|x-1|$ exceeds $\sqrt{25-x^2}$ outside $[-3,4]$, the region above the V-shaped curve and inside the circle exists only for $x\in [-3,4]$:



Since ${\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{3}{5}=\frac{\pi }{2}$ (complementary angles of a 3-4-5 triangle),




Area above the V-curve (the smaller region):


The larger region is the rest of the circle:


So $b=75$, $c=2$, giving $b+c=77$.