To solve the problem step by step, we will first verify that the lines cut the coordinate axes at concyclic points and then find the equation of the circle that passes through these points.
### Step 1: Identify the equations of the lines
The equations of the lines given are:
1. \( 3x - y + 3 = 0 \) (Equation 1)
2. \( x - 3y - 6 = 0 \) (Equation 2)
### Step 2: Find the points where the lines cut the coordinate axes
To find where each line intersects the axes, we will set \( x = 0 \) to find the y-intercept and \( y = 0 \) to find the x-intercept.
**For Equation 1:**
- **Y-intercept:** Set \( x = 0 \):
\[
3(0) - y + 3 = 0 \implies -y + 3 = 0 \implies y = 3
\]
So, the point is \( (0, 3) \).
- **X-intercept:** Set \( y = 0 \):
\[
3x - 0 + 3 = 0 \implies 3x + 3 = 0 \implies 3x = -3 \implies x = -1
\]
So, the point is \( (-1, 0) \).
**For Equation 2:**
- **Y-intercept:** Set \( x = 0 \):
\[
0 - 3y - 6 = 0 \implies -3y - 6 = 0 \implies -3y = 6 \implies y = -2
\]
So, the point is \( (0, -2) \).
- **X-intercept:** Set \( y = 0 \):
\[
x - 3(0) - 6 = 0 \implies x - 6 = 0 \implies x = 6
\]
So, the point is \( (6, 0) \).
### Step 3: List the points of intersection
The points where the lines cut the axes are:
1. \( A(0, 3) \)
2. \( B(-1, 0) \)
3. \( C(0, -2) \)
4. \( D(6, 0) \)
### Step 4: Check if the points are concyclic
To check if points \( A, B, C, D \) are concyclic, we can use the condition:
\[
\text{If } A_1A_2 + A_3A_4 = 0 \text{ or } A_1A_3 \cdot A_2A_4 = A_1A_4 \cdot A_2A_3
\]
Calculating the products:
- \( A_1A_2 = 0 - (-1) + 3 - 0 = 4 \)
- \( A_3A_4 = 0 - 6 + (-2) - 0 = -8 \)
Since \( 4 \cdot (-2) + (-1) \cdot 6 = 0 \), the points are concyclic.
### Step 5: Find the equation of the circle passing through these points
The general equation of a circle is:
\[
x^2 + y^2 + 2gx + 2fy + c = 0
\]
We will substitute the points \( A, B, C, D \) into this equation to find \( g, f, c \).
1. **For point \( A(0, 3) \)**:
\[
0^2 + 3^2 + 2g(0) + 2f(3) + c = 0 \implies 9 + 6f + c = 0 \quad \text{(Equation 1)}
\]
2. **For point \( B(-1, 0) \)**:
\[
(-1)^2 + 0^2 + 2g(-1) + 2f(0) + c = 0 \implies 1 - 2g + c = 0 \quad \text{(Equation 2)}
\]
3. **For point \( C(0, -2) \)**:
\[
0^2 + (-2)^2 + 2g(0) + 2f(-2) + c = 0 \implies 4 - 4f + c = 0 \quad \text{(Equation 3)}
\]
4. **For point \( D(6, 0) \)**:
\[
6^2 + 0^2 + 2g(6) + 2f(0) + c = 0 \implies 36 + 12g + c = 0 \quad \text{(Equation 4)}
\]
### Step 6: Solve the equations
From Equation 1:
\[
c = -9 - 6f
\]
Substituting \( c \) into Equations 2, 3, and 4, we can solve for \( g \) and \( f \) step by step.
After solving these equations, we find:
- \( g = -\frac{5}{2} \)
- \( f = -\frac{1}{2} \)
- \( c = -6 \)
### Step 7: Write the final equation of the circle
Substituting \( g, f, c \) back into the general equation:
\[
x^2 + y^2 - 5x - y - 6 = 0
\]
### Conclusion
Thus, we have shown that the lines cut the coordinate axes at concyclic points and the equation of the circle passing through these points is:
\[
x^2 + y^2 - 5x - y - 6 = 0
\]
