If a circle has two of its diameters along the lines `x+y=5` and `x-y=1` and has area `9 pi` , then the equation of the circle is

Options

A. `x^(2)+y^(2)-6x-4y+4=0`
B. `x^(2)+y^(2)-6x-4y-3=0`
C. `x^(2)+y^(2)-6x-4y-4=0`
D. `x^(2)+y^(2)-6x-4y+3=0`

Explanation

To find the equation of the circle with the given conditions, we will follow these steps:

### Step 1: Find the intersection point of the lines
We have two lines:
1. \( x + y = 5 \) (Equation 1)
2. \( x - y = 1 \) (Equation 2)

To find the center of the circle, we need to find the intersection point of these two lines. We can do this by solving the equations simultaneously.

**Adding Equation 1 and Equation 2:**
\[ (x + y) + (x - y) = 5 + 1 \]
\[ 2x = 6 \implies x = 3 \]

Now, substituting \( x = 3 \) back into Equation 1 to find \( y \):
\[ 3 + y = 5 \implies y = 2 \]

Thus, the center of the circle is at the point \( (3, 2) \).

### Step 2: Find the radius of the circle
We are given that the area of the circle is \( 9\pi \). The formula for the area of a circle is:
\[ \text{Area} = \pi R^2 \]
Setting this equal to \( 9\pi \):
\[ \pi R^2 = 9\pi \]
Dividing both sides by \( \pi \):
\[ R^2 = 9 \implies R = 3 \]

### Step 3: Write the equation of the circle
The standard form of the equation of a circle with center \( (h, k) \) and radius \( R \) is:
\[ (x - h)^2 + (y - k)^2 = R^2 \]
Substituting \( h = 3 \), \( k = 2 \), and \( R = 3 \):
\[ (x - 3)^2 + (y - 2)^2 = 3^2 \]
\[ (x - 3)^2 + (y - 2)^2 = 9 \]

### Step 4: Expand the equation
Now, we can expand the equation:
\[ (x - 3)^2 + (y - 2)^2 = 9 \]
Expanding \( (x - 3)^2 \):
\[ x^2 - 6x + 9 \]
Expanding \( (y - 2)^2 \):
\[ y^2 - 4y + 4 \]
Combining these, we have:
\[ x^2 - 6x + 9 + y^2 - 4y + 4 = 9 \]
Combining like terms:
\[ x^2 + y^2 - 6x - 4y + 13 = 9 \]
Subtracting 9 from both sides:
\[ x^2 + y^2 - 6x - 4y + 4 = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 - 6x - 4y + 4 = 0 \]

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