Animated Solution for Mathematics - Definite Integration: 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

A Geometric Odyssey

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through the intersection of algebra and geometry. We are looking at the area bounded between a circle, $x^2+y^2=25$, and a modulus function, $y=|x-1|$.

This is a classic JEE Advanced problem that tests your ability to visualize, your precision in algebra, and your elegance in calculus. Let us break this down, step by step, and find the beauty hidden within the equations.

The Visualization

Before we touch a single integral sign, we must see the battlefield. We have a circle, $x^2+y^2=25$. This is a perfect, symmetric shape centered at the origin $(0,0)$ with a radius of $5$.

Now, consider the modulus function $y=|x-1|$. This is a V-shaped graph. It turns where the expression inside the modulus is zero, which is at $x=1$. So, the vertex is at $(1,0)$.

Imagine this V-shape sitting on the x-axis, its vertex at $(1,0)$, opening upwards. It slices through our circle. Our goal is to find the area of the 'larger portion' bounded between these two. This means we are looking for the area of the circle minus the 'smaller' region trapped between the circle's arc and the V-shape.

The Intersection Hunt

To find the boundaries of this smaller region, we need to know exactly where the V-shape pierces the circle. The modulus function is piecewise: it is $y=x-1$ when $x\ge 1$ and $y=1-x$ when $x<1$.

Let us tackle the right side first ($x\ge 1$). Substituting $y=x-1$ into the circle equation $x^2+y^2=25$, we get $x^2+(x-1{)}^2=25$. Expanding this, we arrive at $x^2+x^2-2x+1=25$, which simplifies to $2x^2-2x-24=0$, or $x^2-x-12=0$.

Factoring this quadratic gives us $(x-4)(x+3)=0$. Since we are in the region $x\ge 1$, we must reject $x=-3$ and accept $x=4$. Plugging $x=4$ back into $y=x-1$, we find the intersection point $(4,3)$.

Now, for the left side ($x<1$). Substituting $y=1-x$ into the circle equation, we get $x^2+(1-x{)}^2=25$. Because $(1-x{)}^2$ is identical to $(x-1{)}^2$, we land on the same quadratic: $x^2-x-12=0$.

This time, our condition is $x<1$, so we reject $x=4$ and accept $x=-3$. Plugging $x=-3$ into $y=1-x$, we get $y=1-(-3)=4$. Our second intersection point is $(-3,4)$.

The Calculus Strategy

We now have our limits of integration: from $x=-3$ to $x=4$. The smaller area, $A_s$, is the area under the circle minus the area under the modulus function.

Let us handle the circle integral first. The standard integral $\int \sqrt{a^2-x^2}dx$ is $\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}(\frac{x}{a})$. Applying our limits from $-3$ to $4$:

At $x=4$:

At $x=-3$:

Subtracting the lower limit from the upper limit, we get $12+\frac{25}{2}({\sin }^{-1}(\frac{4}{5})+{\sin }^{-1}(\frac{3}{5}))$. Since ${\sin }^{-1}(\frac{4}{5})+{\sin }^{-1}(\frac{3}{5})=\frac{\pi }{2}$, the area under the circle is $12+\frac{25\pi }{4}$.

The Geometric Shortcut

Now, for the modulus part. Do not waste time integrating $|x-1|$. Look at the graph.

From $x=-3$ to $x=1$, we have a triangle with base $4$ and height $4$. Area = $\frac{1}{2}\times 4\times 4=8$.

From $x=1$ to $x=4$, we have a triangle with base $3$ and height $3$. Area = $\frac{1}{2}\times 3\times 3=4.5$.

Total area = $12.5=\frac{25}{2}$.

The Final Synthesis

Subtracting the modulus area from the circle area:

Finally, the larger area $A_L$ is the total area of the circle ($25\pi$) minus $A_s$:

Comparing this to $\frac{1}{4}(b\pi +c)$, we find $b=75$ and $c=2$. The sum $b+c=77$.

There you have it. A perfect blend of coordinate geometry, trigonometric identities, and calculus. You have conquered the problem!