To find the center of the circle passing through the points of intersection of the curves \((2x + 3y + 4)(3x + 2y - 1) = 0\) and \(xy = 0\), we will follow these steps:
### Step 1: Identify the curves
The first curve is given by the equation:
\[
(2x + 3y + 4)(3x + 2y - 1) = 0
\]
This equation represents two lines:
1. \(2x + 3y + 4 = 0\)
2. \(3x + 2y - 1 = 0\)
The second curve is:
\[
xy = 0
\]
This means that either \(x = 0\) or \(y = 0\), which represents the x-axis and y-axis.
### Step 2: Find points of intersection
We will find the points of intersection of the lines with the axes.
**For \(x = 0\):**
1. Substitute \(x = 0\) into \(2x + 3y + 4 = 0\):
\[
3y + 4 = 0 \implies y = -\frac{4}{3}
\]
So, the point is \((0, -\frac{4}{3})\).
2. Substitute \(x = 0\) into \(3x + 2y - 1 = 0\):
\[
2y - 1 = 0 \implies y = \frac{1}{2}
\]
So, the point is \((0, \frac{1}{2})\).
**For \(y = 0\):**
1. Substitute \(y = 0\) into \(2x + 3y + 4 = 0\):
\[
2x + 4 = 0 \implies x = -2
\]
So, the point is \((-2, 0)\).
2. Substitute \(y = 0\) into \(3x + 2y - 1 = 0\):
\[
3x - 1 = 0 \implies x = \frac{1}{3}
\]
So, the point is \((\frac{1}{3}, 0)\).
### Step 3: List the points of intersection
The points of intersection are:
1. \((0, -\frac{4}{3})\)
2. \((0, \frac{1}{2})\)
3. \((-2, 0)\)
4. \((\frac{1}{3}, 0)\)
### Step 4: Find the equation of the circle
The center of the circle passing through these points can be found using the general form of the circle equation:
\[
x^2 + y^2 + 2gx + 2fy + c = 0
\]
We need to find the values of \(g\), \(f\), and \(c\).
### Step 5: Substitute the points into the circle equation
We will substitute the points into the circle equation to form a system of equations.
1. For \((0, -\frac{4}{3})\):
\[
0^2 + \left(-\frac{4}{3}\right)^2 + 2g(0) + 2f\left(-\frac{4}{3}\right) + c = 0
\]
\[
\frac{16}{9} - \frac{8f}{3} + c = 0 \quad \text{(Equation 1)}
\]
2. For \((0, \frac{1}{2})\):
\[
0^2 + \left(\frac{1}{2}\right)^2 + 2g(0) + 2f\left(\frac{1}{2}\right) + c = 0
\]
\[
\frac{1}{4} + f + c = 0 \quad \text{(Equation 2)}
\]
3. For \((-2, 0)\):
\[
(-2)^2 + 0^2 + 2g(-2) + 2f(0) + c = 0
\]
\[
4 - 4g + c = 0 \quad \text{(Equation 3)}
\]
4. For \((\frac{1}{3}, 0)\):
\[
\left(\frac{1}{3}\right)^2 + 0^2 + 2g\left(\frac{1}{3}\right) + 2f(0) + c = 0
\]
\[
\frac{1}{9} + \frac{2g}{3} + c = 0 \quad \text{(Equation 4)}
\]
### Step 6: Solve the equations
Now, we will solve these equations to find \(g\), \(f\), and \(c\).
From Equation 1:
\[
c = \frac{8f}{3} - \frac{16}{9}
\]
From Equation 2:
\[
c = -f - \frac{1}{4}
\]
Setting the two expressions for \(c\) equal:
\[
\frac{8f}{3} - \frac{16}{9} = -f - \frac{1}{4}
\]
Multiply through by 36 to eliminate fractions:
\[
96f - 64 = -36f - 9
\]
\[
132f = 55 \implies f = \frac{55}{132} = \frac{5}{12}
\]
Now substitute \(f\) back to find \(c\):
\[
c = -\frac{5}{12} - \frac{1}{4} = -\frac{5}{12} - \frac{3}{12} = -\frac{8}{12} = -\frac{2}{3}
\]
Now substitute \(f\) into Equation 3 to find \(g\):
\[
4 - 4g - \frac{2}{3} = 0 \implies 4 - \frac{2}{3} = 4g \implies \frac{12}{3} - \frac{2}{3} = 4g \implies \frac{10}{3} = 4g \implies g = \frac{10}{12} = \frac{5}{6}
\]
### Step 7: Find the center of the circle
The center of the circle is given by the coordinates \((-g, -f)\):
\[
\text{Center} = \left(-\frac{5}{6}, -\frac{5}{12}\right)
\]
### Final Answer
The center of the circle passing through the points of intersection of the curves is:
\[
\left(-\frac{5}{6}, -\frac{5}{12}\right)
\]
