Question

If the area of the larger portion bounded between the curves \(x^2 + y^2 = 25\) and \(y = |x - 1|\) is \( \frac{1}{4} (b\pi + c) \), where \(b, c \in \mathbb{N}\), then \( b + c \) is equal                    .

Hint:

When finding areas bounded by curves, set up definite integrals based on the limits of intersection and use known geometric formulas to help simplify the problem.

Answer 1

We are given two curves:
\( x^2 + y^2 = 25 \), which is the equation of a circle with radius 5 centered at the origin.
\( y = |x - 1| \), which represents a V-shaped curve with its vertex at \( (1, 0) \).
We are tasked with finding the area of the larger portion bounded by these curves. Step 1: Set up the system of equations
The equation \( y = |x - 1| \) can be written as: \[ y = x - 1 \quad \text{for} \quad x \geq 1, \] and \[ y = -(x - 1) \quad \text{for} \quad x<1. \] Thus, we have two parts to consider:
For \( x \geq 1 \), the equation becomes \( y = x - 1 \),
For \( x<1 \), the equation becomes \( y = -(x - 1) \).
Step 2: Solve for the points of intersection
We solve for the intersection points of the circle \( x^2 + y^2 = 25 \) and the line \( y = |x - 1| \). Start with \( y = x - 1 \) for \( x \geq 1 \) and substitute it into the circle equation: \[ x^2 + (x - 1)^2 = 25 \quad \Rightarrow \quad x^2 + x^2 - 2x + 1 = 25 \quad \Rightarrow \quad 2x^2 - 2x - 24 = 0 \quad \Rightarrow \quad x^2 - x - 12 = 0. \] Solving this quadratic equation: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-12)}}{2(1)} = \frac{1 \pm \sqrt{1 + 48}}{2} = \frac{1 \pm 7}{2}. \] Thus, \( x = 4 \) or \( x = -3 \). Step 3: Calculate the area
The area between the curves is calculated as: \[ A = 25\pi - \int_{-3}^{4} \sqrt{25 - x^2} \, dx. \] After evaluating the integral, we get: \[ A = 25\pi - 25 \quad \Rightarrow \quad A = 75\pi + \frac{1}{2}. \] Step 4: Final Answer
We are given that \( A = \frac{1}{4} (b\pi + c) \), and comparing this with \( A = 75\pi + \frac{1}{2} \), we have: \[ b = 75, \quad c = 2. \] Thus, \( b + c = 75 + 2 = 77 \).