What is the area bounded by `x^(2)+y^(2)-8x-6y=25,xge4` and `yge3`?

Options

A. 39.3
B. 78.6
C. 123.4
D. 157

Answer

To find the area bounded by the equation \(x^2 + y^2 - 8x - 6y = 25\) along with the constraints \(x \geq 4\) and \(y \geq 3\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
We start with the given equation:
\[ x^2 + y^2 - 8x - 6y = 25 \]
We will complete the square for both \(x\) and \(y\).

### Step 2: Completing the Square
1. For \(x\):
\[ x^2 - 8x \quad \text{(take half of -8, square it: } (-4)^2 = 16\text{)} \]
So, we rewrite it as:
\[ (x - 4)^2 - 16 \]

2. For \(y\):
\[ y^2 - 6y \quad \text{(take half of -6, square it: } (-3)^2 = 9\text{)} \]
So, we rewrite it as:
\[ (y - 3)^2 - 9 \]

Putting it all together:
\[ (x - 4)^2 - 16 + (y - 3)^2 - 9 = 25 \]
This simplifies to:
\[ (x - 4)^2 + (y - 3)^2 = 50 \]

### Step 3: Identify the Circle's Center and Radius
From the equation \((x - 4)^2 + (y - 3)^2 = 50\), we can identify:
- Center: \((4, 3)\)
- Radius: \(r = \sqrt{50} = 5\sqrt{2}\)

### Step 4: Determine the Area of the Sector
The area we are interested in is the sector of the circle that is bounded by the lines \(x = 4\) and \(y = 3\). This sector corresponds to a quarter of the circle (90 degrees).

1. The area of the entire circle is:
\[ \text{Area} = \pi r^2 = \pi (50) = 50\pi \]

2. The area of the quarter circle (90 degrees) is:
\[ \text{Area of sector} = \frac{1}{4} \times 50\pi = \frac{50\pi}{4} = \frac{25\pi}{2} \]

### Step 5: Approximate the Area
Using \(\pi \approx 3.14\):
\[ \text{Area} \approx \frac{25 \times 3.14}{2} = \frac{78.5}{2} = 39.25 \]

### Final Answer
Thus, the area bounded by the given equations is approximately:
\[ \text{Area} \approx 39.25 \]