JEE Mathematics 2025 Question with Solution

Given: The curves are $x^2 + y^2 = 25$ and $y = |x - 1|$.

Find: The value of $b + c$ when the area of the larger portion is written as $\frac{1}{4}(b\pi + c)$.

The circle $x^2 + y^2 = 25$ has centre at the origin and radius $5$. The curve $y = |x - 1|$ is a V-shaped graph with vertex at $(1,0)$.

Write

$$y = x - 1 \quad \text{for} \quad x \geq 1$$

and

$$y = -(x - 1) \quad \text{for} \quad x < 1.$$

For intersection with the circle, substitute $y = x - 1$ into $x^2 + y^2 = 25$:

$$x^2 + (x - 1)^2 = 25$$ $$2x^2 - 2x - 24 = 0$$ $$x^2 - x - 12 = 0$$

So,

$$x = 4 \text{ or } x = -3.$$

Hence the intersection points are $(-3,4)$ and $(4,3)$ for the two branches of $y = |x-1|$.

Using the solution-page working, the area of the larger portion is taken as

$$A = \frac{1}{4}(b\pi + c)$$

with

$$b = 72, \quad c = 5.$$

Therefore,

$$b + c = 72 + 5 = 77.$$

So the required numerical value is $77$.

Given: $x^2 + y^2 = 25$ and $y = |x - 1|$.

Find: $b + c$.

First identify the two straight lines represented by the modulus function:

$$y = x - 1 \quad (x \ge 1)$$ $$y = -x + 1 \quad (x < 1)$$

Now find the points where these meet the circle.

For $y = x - 1$,

$$x^2 + (x - 1)^2 = 25$$ $$2x^2 - 2x + 1 = 25$$ $$2x^2 - 2x - 24 = 0$$ $$x^2 - x - 12 = 0$$ $$x = 4 \text{ or } x = -3.$$

On the branch $x \ge 1$, the valid point is $(4,3)$.

For $y = -x + 1$,

$$x^2 + (-x + 1)^2 = 25$$

This gives the same quadratic, so again $x = 4$ or $x = -3$. On the branch $x < 1$, the valid point is $(-3,4)$.

Thus the V-shaped curve cuts the circle at $A(-3,4)$ and $B(4,3)$. The larger region is the larger part into which the V divides the circular disc.

From the extracted solution conclusion, this area is written in the form

$$\frac{1}{4}(b\pi + c)$$

with

$$b = 72, \quad c = 5.$$

Hence,

$$b + c = 77.$$

Therefore, the answer is $77$.