If the lines `2x+3y+1=0` and `3x-y-4=0` lie along diameters of a circle of circumference `10 pi`, then the equation of the circle is

Options

A. `x^(2)+y^(2)+2x-2y-23=0`
B. `x^(2)+y^(2)-2x-2y-23=0`
C. `x^(2)+y^(2)+2x+2y-23=0`
D. `x^(2)+y^(2)-2x+2y-23=0`

Explanation

To find the equation of the circle given that the lines \(2x + 3y + 1 = 0\) and \(3x - y - 4 = 0\) lie along the diameters of the circle with a circumference of \(10\pi\), we can follow these steps:

### Step 1: Find the intersection point of the two lines
We have the equations of the lines:
1. \(2x + 3y + 1 = 0\)
2. \(3x - y - 4 = 0\)

To find the intersection point, we can express \(y\) in terms of \(x\) from the second equation:
\[ y = 3x - 4 \]

Now, substitute this expression for \(y\) into the first equation:
\[ 2x + 3(3x - 4) + 1 = 0 \]
\[ 2x + 9x - 12 + 1 = 0 \]
\[ 11x - 11 = 0 \]
\[ 11x = 11 \implies x = 1 \]

Now, substitute \(x = 1\) back into the equation for \(y\):
\[ y = 3(1) - 4 = 3 - 4 = -1 \]

Thus, the intersection point (which is the center of the circle) is:
\[ (1, -1) \]

### Step 2: Find the radius of the circle
We know the circumference \(C\) of the circle is given by:
\[ C = 2\pi r \]
Given \(C = 10\pi\), we can set up the equation:
\[ 2\pi r = 10\pi \]

Dividing both sides by \(2\pi\):
\[ r = \frac{10\pi}{2\pi} = 5 \]

### 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 \(h = 1\), \(k = -1\), and \(r = 5\):
\[ (x - 1)^2 + (y + 1)^2 = 5^2 \]
\[ (x - 1)^2 + (y + 1)^2 = 25 \]

### Step 4: Expand the equation
Now, we can expand the equation:
\[ (x - 1)^2 + (y + 1)^2 = 25 \]
\[ x^2 - 2x + 1 + y^2 + 2y + 1 = 25 \]
\[ x^2 + y^2 - 2x + 2y + 2 = 25 \]

Now, rearranging gives:
\[ x^2 + y^2 - 2x + 2y - 23 = 0 \]

### Final Answer
Thus, the equation of the circle is:
\[ x^2 + y^2 - 2x + 2y - 23 = 0 \]