Let `A={(x,y)in R^(2):yge0,2xleylesqrt(4-(x-1)^(2)})` and `B={(x,y)in R times R:0leylemin{2x,sqrt(4-(x-1)^(2)})}`
Then the ratio of the area of `A` to the area of `B` is

Options

A. `(pi)/(pi+1)`
B. `(pi-1)/(pi+1)`
C. `(pi+1)/(pi-1)`
D. `(pi)/(pi-1),`

Answer

To solve the problem, we need to find the areas of the sets \( A \) and \( B \) defined in the question and then calculate the ratio of these areas.

### Step 1: Understanding the set \( A \)

The set \( A \) is defined as:
\[ A = \{(x, y) \in \mathbb{R}^2 : y \geq 0, 2x \leq y \leq \sqrt{4 - (x - 1)^2}\} \]

1. **Identify the boundaries**:
- The line \( y = 2x \) is a straight line that passes through the origin and has a slope of 2.
- The equation \( y = \sqrt{4 - (x - 1)^2} \) describes the upper half of a circle centered at \( (1, 0) \) with a radius of 2.

2. **Finding intersection points**:
- To find the intersection of \( y = 2x \) and \( y = \sqrt{4 - (x - 1)^2} \), we set:
\[ 2x = \sqrt{4 - (x - 1)^2} \]
- Squaring both sides:
\[ 4x^2 = 4 - (x - 1)^2 \]
- Expanding and rearranging gives:
\[ 4x^2 = 4 - (x^2 - 2x + 1) \implies 4x^2 = 3 + 2x - x^2 \implies 5x^2 - 2x - 3 = 0 \]
- Solving this quadratic using the quadratic formula:
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 5 \cdot (-3)}}{2 \cdot 5} = \frac{2 \pm \sqrt{4 + 60}}{10} = \frac{2 \pm 8}{10} \]
- This gives \( x = 1 \) and \( x = -\frac{3}{5} \).

3. **Finding corresponding \( y \) values**:
- For \( x = 1 \): \( y = 2(1) = 2 \)
- For \( x = -\frac{3}{5} \): \( y = 2(-\frac{3}{5}) = -\frac{6}{5} \) (not valid since \( y \geq 0 \))

Thus, the relevant points are \( (1, 2) \) and \( (0, 0) \).

### Step 2: Area of set \( A \)

The area of the region \( A \) can be calculated as the area under the curve \( y = \sqrt{4 - (x - 1)^2} \) from \( x = 0 \) to \( x = 1 \) minus the area of the triangle formed by the line \( y = 2x \) and the x-axis.

1. **Area under the curve**:
- The area of the quarter circle from \( x = 0 \) to \( x = 2 \) is:
\[ \text{Area of quarter circle} = \frac{1}{4} \pi r^2 = \frac{1}{4} \pi (2^2) = \pi \]

2. **Area of triangle**:
- The triangle has a base of 1 and a height of 2:
\[ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 2 = 1 \]

3. **Total area of \( A \)**:
\[ \text{Area of } A = \pi - 1 \]

### Step 3: Understanding the set \( B \)

The set \( B \) is defined as:
\[ B = \{(x, y) \in \mathbb{R} \times \mathbb{R} : 0 \leq y \leq \min(2x, \sqrt{4 - (x - 1)^2})\} \]

1. **Finding the area of set \( B \)**:
- The area is bounded by the same curves as in \( A \) but extends further since \( y \) can go up to \( 2x \) for \( x \geq 0 \) and \( y \) can also go up to the circle until \( x = 3 \).

2. **Area calculation**:
- The area under \( y = 2x \) from \( x = 0 \) to \( x = 1.5 \) (where \( 2x \) intersects the circle):
\[ \text{Area under } y = 2x = \int_0^{1.5} 2x \, dx = [x^2]_0^{1.5} = (1.5)^2 = 2.25 \]
- The area under the circle from \( x = 1.5 \) to \( x = 3 \):
\[ \text{Area under circle} = \text{Area of quarter circle} - \text{Area of triangle} \]
- The total area of \( B \) can be calculated similarly, leading to:
\[ \text{Area of } B = 2 + \frac{\pi}{4} - 1 = 1 + \frac{\pi}{4} \]

### Step 4: Ratio of areas

Now, we can find the ratio of the areas:
\[ \text{Ratio} = \frac{\text{Area of } A}{\text{Area of } B} = \frac{\pi - 1}{1 + \frac{\pi}{4}} \]

### Final Result

After simplifying, we find the ratio of the area of \( A \) to the area of \( B \).