If the lines `x+y+3=0 and 2x-y-9=0` lie along diameters of a circle of circumference `8pi` then the equation of the circle is :

Options

A. `x^(2)+y^(2)-4x+10y+13=0`
B. `x^(2)+y^(2)+4x-10y+13=0`
C. `x^(2)+y^(2)-4x-10y+13=0`
D. `x^(2)+y^(2)+4x+10y+13=0`

Explanation

To find the equation of the circle given that the lines \(x + y + 3 = 0\) and \(2x - y - 9 = 0\) lie along diameters of the circle with a circumference of \(8\pi\), we can follow these steps:

### Step 1: Find the center of the circle
The center of the circle lies at the intersection of the two lines. We will solve the equations of the lines simultaneously.

1. **Equation of Line 1**:
\[ x + y + 3 = 0 \implies y = -x - 3 \]

2. **Equation of Line 2**:
\[ 2x - y - 9 = 0 \implies y = 2x - 9 \]

Now, we set the two expressions for \(y\) equal to each other:
\[ -x - 3 = 2x - 9 \]

### Step 2: Solve for \(x\)
Rearranging the equation:
\[ -x - 2x = -9 + 3 \]
\[ -3x = -6 \implies x = 2 \]

### Step 3: Solve for \(y\)
Now substitute \(x = 2\) back into one of the equations to find \(y\):
\[ y = -2 - 3 = -5 \]

Thus, the center \(C\) of the circle is at the point \((2, -5)\).

### Step 4: Find the radius of the circle
The circumference \(C\) of the circle is given as \(8\pi\). The formula for the circumference of a circle is:
\[ C = 2\pi r \]
Setting this equal to \(8\pi\):
\[ 2\pi r = 8\pi \]
Dividing both sides by \(2\pi\):
\[ r = 4 \]

### Step 5: Write the equation of the circle
The standard form of the equation of a circle is:
\[ (x - x_1)^2 + (y - y_1)^2 = r^2 \]
Where \((x_1, y_1)\) is the center and \(r\) is the radius. Substituting the values we found:
\[ (x - 2)^2 + (y + 5)^2 = 4^2 \]
This simplifies to:
\[ (x - 2)^2 + (y + 5)^2 = 16 \]

### Step 6: Expand the equation
Expanding the left side:
\[ (x^2 - 4x + 4) + (y^2 + 10y + 25) = 16 \]
Combining like terms:
\[ x^2 + y^2 - 4x + 10y + 29 = 16 \]
Rearranging gives:
\[ x^2 + y^2 - 4x + 10y + 13 = 0 \]

Thus, the equation of the circle is:
\[ \boxed{x^2 + y^2 - 4x + 10y + 13 = 0} \]