The equation of the circle circumscribing the triangle formed by the lines `x+y = 6,2x + y = 4` and `x + 2y = 5` is

Options

A. `x^(2) +y^(2) +17x +19y -50 = 0`
B. `x^(2) +y^(2) -17x -19y -50 = 0`
C. `x^(2) +y^(2) +17x -19y -50 = 0`
D. `x^(2) +y^(2) -17x -19y +50 = 0`

Answer

To find the equation of the circle circumscribing the triangle formed by the lines \(x + y = 6\), \(2x + y = 4\), and \(x + 2y = 5\), we can follow these steps:

### Step 1: Find the points of intersection of the lines
We need to find the vertices of the triangle formed by the intersection of the three lines.

1. **Intersection of \(x + y = 6\) and \(2x + y = 4\)**:
\[ \begin{align*} x + y &= 6 \quad \text{(1)} \\ 2x + y &= 4 \quad \text{(2)} \end{align*} \]
Subtract (1) from (2):
\[ (2x + y) - (x + y) = 4 - 6 \\ x = -2 \]
Substitute \(x = -2\) into (1):
\[ -2 + y = 6 \\ y = 8 \]
So, point \(A = (-2, 8)\).

2. **Intersection of \(2x + y = 4\) and \(x + 2y = 5\)**:
\[ \begin{align*} 2x + y &= 4 \quad \text{(3)} \\ x + 2y &= 5 \quad \text{(4)} \end{align*} \]
From (3), express \(y\):
\[ y = 4 - 2x \]
Substitute into (4):
\[ x + 2(4 - 2x) = 5 \\ x + 8 - 4x = 5 \\ -3x + 8 = 5 \\ -3x = -3 \\ x = 1 \]
Substitute \(x = 1\) into (3):
\[ 2(1) + y = 4 \\ y = 2 \]
So, point \(B = (1, 2)\).

3. **Intersection of \(x + y = 6\) and \(x + 2y = 5\)**:
\[ \begin{align*} x + y &= 6 \quad \text{(5)} \\ x + 2y &= 5 \quad \text{(6)} \end{align*} \]
From (5), express \(y\):
\[ y = 6 - x \]
Substitute into (6):
\[ x + 2(6 - x) = 5 \\ x + 12 - 2x = 5 \\ -x + 12 = 5 \\ -x = -7 \\ x = 7 \]
Substitute \(x = 7\) into (5):
\[ 7 + y = 6 \\ y = -1 \]
So, point \(C = (7, -1)\).

### Step 2: Find the circumcenter of the triangle
The circumcenter is the point where the perpendicular bisectors of the sides of the triangle intersect.

1. **Midpoint of \(AB\)**:
\[ M_{AB} = \left(\frac{-2 + 1}{2}, \frac{8 + 2}{2}\right) = \left(-\frac{1}{2}, 5\right) \]
The slope of \(AB\) is:
\[ m_{AB} = \frac{2 - 8}{1 - (-2)} = \frac{-6}{3} = -2 \]
The slope of the perpendicular bisector is:
\[ m_{pAB} = \frac{1}{2} \]
The equation of the perpendicular bisector:
\[ y - 5 = \frac{1}{2}\left(x + \frac{1}{2}\right) \\ 2y - 10 = x + \frac{1}{2} \\ x - 2y + 10.5 = 0 \quad \text{(7)} \]

2. **Midpoint of \(BC\)**:
\[ M_{BC} = \left(\frac{1 + 7}{2}, \frac{2 + (-1)}{2}\right) = \left(4, \frac{1}{2}\right) \]
The slope of \(BC\) is:
\[ m_{BC} = \frac{-1 - 2}{7 - 1} = \frac{-3}{6} = -\frac{1}{2} \]
The slope of the perpendicular bisector is:
\[ m_{pBC} = 2 \]
The equation of the perpendicular bisector:
\[ y - \frac{1}{2} = 2(x - 4) \\ y - \frac{1}{2} = 2x - 8 \\ 2x - y - \frac{15}{2} = 0 \quad \text{(8)} \]

### Step 3: Solve equations (7) and (8) to find the circumcenter
To find the intersection of lines (7) and (8):
1. From (7): \(x - 2y + 10.5 = 0 \Rightarrow x = 2y - 10.5\).
2. Substitute into (8):
\[ 2(2y - 10.5) - y - \frac{15}{2} = 0 \\ 4y - 21 - y - \frac{15}{2} = 0 \\ 3y - 21 - 7.5 = 0 \\ 3y = 28.5 \\ y = 9.5 \]
Substitute back to find \(x\):
\[ x = 2(9.5) - 10.5 = 8.5 \]
So, the circumcenter is \((8.5, 9.5)\).

### Step 4: Find the radius
Using the distance formula, the radius \(r\) is the distance from the circumcenter to any vertex, say \(A\):
\[ r = \sqrt{(8.5 - (-2))^2 + (9.5 - 8)^2} = \sqrt{(10.5)^2 + (1.5)^2} = \sqrt{110.25 + 2.25} = \sqrt{112.5} \]

### Step 5: Write the equation of the circle
The equation of the circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius.
\[ (x - 8.5)^2 + (y - 9.5)^2 = 112.5 \]

### Final Equation
Expanding this gives:
\[ x^2 - 17x + 72.25 + y^2 - 19y + 90.25 = 112.5 \\ x^2 + y^2 - 17x - 19y + 50 = 0 \]

### Conclusion
The equation of the circle circumscribing the triangle formed by the given lines is:
\[ x^2 + y^2 - 17x - 19y + 50 = 0 \]