To find the equation of the circle passing through the vertices of the triangle formed by the lines \(x - y = 0\), \(3x + 2y = 5\), and \(x + 2y = 5\), we will follow these steps:
### Step 1: Find the vertices of the triangle
1. **Find the intersection of the lines \(x - y = 0\) and \(3x + 2y = 5\)**:
- From \(x - y = 0\), we have \(y = x\).
- Substitute \(y = x\) into \(3x + 2y = 5\):
\[
3x + 2x = 5 \implies 5x = 5 \implies x = 1 \implies y = 1
\]
- Thus, the first vertex \(A\) is \((1, 1)\).
2. **Find the intersection of the lines \(3x + 2y = 5\) and \(x + 2y = 5\)**:
- From \(x + 2y = 5\), we can express \(x\) as \(x = 5 - 2y\).
- Substitute \(x = 5 - 2y\) into \(3x + 2y = 5\):
\[
3(5 - 2y) + 2y = 5 \implies 15 - 6y + 2y = 5 \implies -4y = -10 \implies y = \frac{5}{2}
\]
- Substitute \(y = \frac{5}{2}\) back into \(x + 2y = 5\):
\[
x + 5 = 5 \implies x = 0
\]
- Thus, the second vertex \(B\) is \((0, \frac{5}{2})\).
3. **Find the intersection of the lines \(x - y = 0\) and \(x + 2y = 5\)**:
- From \(x - y = 0\), we have \(y = x\).
- Substitute \(y = x\) into \(x + 2y = 5\):
\[
x + 2x = 5 \implies 3x = 5 \implies x = \frac{5}{3} \implies y = \frac{5}{3}
\]
- Thus, the third vertex \(C\) is \((\frac{5}{3}, \frac{5}{3})\).
### Step 2: Use the vertices to find the equation of the circle
The vertices of the triangle are:
- \(A(1, 1)\)
- \(B(0, \frac{5}{2})\)
- \(C(\frac{5}{3}, \frac{5}{3})\)
The general equation of a circle is given by:
\[
x^2 + y^2 + 2gx + 2fy + c = 0
\]
We will substitute the coordinates of the vertices into this equation to form a system of equations.
1. **Substituting vertex \(A(1, 1)\)**:
\[
1^2 + 1^2 + 2g(1) + 2f(1) + c = 0 \implies 2 + 2g + 2f + c = 0 \implies 2g + 2f + c = -2 \quad \text{(Equation 1)}
\]
2. **Substituting vertex \(B(0, \frac{5}{2})\)**:
\[
0^2 + \left(\frac{5}{2}\right)^2 + 2g(0) + 2f\left(\frac{5}{2}\right) + c = 0 \implies \frac{25}{4} + 5f + c = 0 \implies 5f + c = -\frac{25}{4} \quad \text{(Equation 2)}
\]
3. **Substituting vertex \(C(\frac{5}{3}, \frac{5}{3})\)**:
\[
\left(\frac{5}{3}\right)^2 + \left(\frac{5}{3}\right)^2 + 2g\left(\frac{5}{3}\right) + 2f\left(\frac{5}{3}\right) + c = 0
\]
\[
\frac{25}{9} + \frac{25}{9} + \frac{10g}{3} + \frac{10f}{3} + c = 0 \implies \frac{50}{9} + \frac{10g}{3} + \frac{10f}{3} + c = 0 \quad \text{(Equation 3)}
\]
### Step 3: Solve the system of equations
1. From Equation 1:
\[
c = -2 - 2g - 2f
\]
2. Substitute \(c\) into Equation 2:
\[
5f + (-2 - 2g - 2f) = -\frac{25}{4} \implies 3f - 2 - 2g = -\frac{25}{4}
\]
\[
3f - 2g = -\frac{25}{4} + 2 = -\frac{25}{4} + \frac{8}{4} = -\frac{17}{4} \quad \text{(Equation 4)}
\]
3. Substitute \(c\) into Equation 3:
\[
\frac{50}{9} + \frac{10g}{3} + \frac{10f}{3} + (-2 - 2g - 2f) = 0
\]
\[
\frac{50}{9} - 2 + \frac{10g}{3} - 2g + \frac{10f}{3} - 2f = 0
\]
\[
\frac{50}{9} - \frac{18}{9} + \frac{10g - 6g + 10f - 6f}{3} = 0 \implies \frac{32}{9} + \frac{4g + 4f}{3} = 0
\]
\[
4g + 4f = -\frac{32}{9} \implies g + f = -\frac{8}{9} \quad \text{(Equation 5)}
\]
### Step 4: Solve Equations 4 and 5
From Equation 5:
\[
f = -\frac{8}{9} - g
\]
Substituting \(f\) into Equation 4:
\[
3\left(-\frac{8}{9} - g\right) - 2g = -\frac{17}{4}
\]
\[
-\frac{24}{9} - 3g - 2g = -\frac{17}{4}
\]
\[
-\frac{24}{9} - 5g = -\frac{17}{4}
\]
To solve for \(g\), we can convert \(-\frac{24}{9}\) and \(-\frac{17}{4}\) to a common denominator:
\[
-\frac{24}{9} = -\frac{96}{36}, \quad -\frac{17}{4} = -\frac{153}{36}
\]
\[
-5g = -\frac{153}{36} + \frac{96}{36} \implies -5g = -\frac{57}{36} \implies g = \frac{57}{180} = \frac{19}{60}
\]
Now substitute \(g\) back to find \(f\):
\[
f = -\frac{8}{9} - \frac{19}{60}
\]
Finding a common denominator:
\[
f = -\frac{480}{540} - \frac{171}{540} = -\frac{651}{540} = -\frac{217}{180}
\]
Finally, substitute \(g\) and \(f\) back to find \(c\):
\[
c = -2 - 2\left(\frac{19}{60}\right) - 2\left(-\frac{217}{180}\right)
\]
Calculating \(c\):
\[
c = -2 - \frac{38}{60} + \frac{434}{180}
\]
Finding a common denominator:
\[
c = -\frac{360}{180} - \frac{114}{180} + \frac{434}{180} = \frac{-360 - 114 + 434}{180} = \frac{-40}{180} = -\frac{2}{9}
\]
### Final Equation of the Circle
Now substituting \(g\), \(f\), and \(c\) into the general circle equation:
\[
x^2 + y^2 + 2\left(\frac{19}{60}\right)x + 2\left(-\frac{217}{180}\right)y - \frac{2}{9} = 0
\]
After simplifying, we can express the equation in standard form. The final equation of the circle is:
\[
\text{Circle: } 60x^2 + 60y^2 + 19x - 217y - 40 = 0
\]
