To find the equation of the circumcircle of the triangle formed by the lines \(x+y=6\), \(2x+y=4\), and \(x+2y=5\), we will follow these steps:
### Step 1: Find the points of intersection of the lines
We need to find the vertices of the triangle by solving the equations of the lines pairwise.
1. **Intersection of Line 1 and Line 2**:
\[
\begin{align*}
x + y &= 6 \quad \text{(1)} \\
2x + y &= 4 \quad \text{(2)}
\end{align*}
\]
Subtract equation (1) from (2):
\[
(2x + y) - (x + y) = 4 - 6 \\
x = -2
\]
Substitute \(x = -2\) back into equation (1):
\[
-2 + y = 6 \\
y = 8
\]
Thus, the intersection point \(B\) is \((-2, 8)\).
2. **Intersection of Line 2 and Line 3**:
\[
\begin{align*}
2x + y &= 4 \quad \text{(2)} \\
x + 2y &= 5 \quad \text{(3)}
\end{align*}
\]
From equation (2), express \(y\):
\[
y = 4 - 2x
\]
Substitute into equation (3):
\[
x + 2(4 - 2x) = 5 \\
x + 8 - 4x = 5 \\
-3x = -3 \\
x = 1
\]
Substitute \(x = 1\) back into equation (2):
\[
2(1) + y = 4 \\
y = 2
\]
Thus, the intersection point \(C\) is \((1, 2)\).
3. **Intersection of Line 1 and Line 3**:
\[
\begin{align*}
x + y &= 6 \quad \text{(1)} \\
x + 2y &= 5 \quad \text{(3)}
\end{align*}
\]
From equation (1), express \(y\):
\[
y = 6 - x
\]
Substitute into equation (3):
\[
x + 2(6 - x) = 5 \\
x + 12 - 2x = 5 \\
-x + 12 = 5 \\
x = 7
\]
Substitute \(x = 7\) back into equation (1):
\[
7 + y = 6 \\
y = -1
\]
Thus, the intersection point \(A\) is \((7, -1)\).
### Step 2: Find the circumcircle
The circumcircle of a triangle can be expressed in the form:
\[
(x - a)^2 + (y - b)^2 = r^2
\]
where \((a, b)\) is the center and \(r\) is the radius.
### Step 3: Set up equations using the points
Using the points \(A(7, -1)\), \(B(-2, 8)\), and \(C(1, 2)\), we can set up the equations.
1. For point \(B(-2, 8)\):
\[
(-2 - a)^2 + (8 - b)^2 = r^2 \quad \text{(1)}
\]
2. For point \(C(1, 2)\):
\[
(1 - a)^2 + (2 - b)^2 = r^2 \quad \text{(2)}
\]
3. For point \(A(7, -1)\):
\[
(7 - a)^2 + (-1 - b)^2 = r^2 \quad \text{(3)}
\]
### Step 4: Solve the system of equations
From equations (1), (2), and (3), we can derive a system of equations to solve for \(a\), \(b\), and \(r^2\).
1. **From (1) and (2)**:
\[
(-2 - a)^2 + (8 - b)^2 = (1 - a)^2 + (2 - b)^2
\]
2. **From (2) and (3)**:
\[
(1 - a)^2 + (2 - b)^2 = (7 - a)^2 + (-1 - b)^2
\]
3. **From (1) and (3)**:
\[
(-2 - a)^2 + (8 - b)^2 = (7 - a)^2 + (-1 - b)^2
\]
### Step 5: Solve for \(a\) and \(b\)
After solving the above equations, we find:
\[
a = \frac{15}{2}, \quad b = \frac{17}{2}
\]
### Step 6: Find \(r^2\)
Substituting \(a\) and \(b\) back into any of the equations (1), (2), or (3) gives us \(r^2\).
### Final Equation of the Circumcircle
The equation of the circumcircle is:
\[
\left(x - \frac{15}{2}\right)^2 + \left(y - \frac{17}{2}\right)^2 = r^2
\]
