To solve the problem, we need to find the value of \( \lambda \) such that a circle passes through the points of intersection of the lines \( 2x - y + 1 = 0 \) and \( x + \lambda y - 3 = 0 \) with the axes of reference.
### Step 1: Find the points of intersection of the lines with the axes.
1. **For the line \( 2x - y + 1 = 0 \)**:
- To find the x-intercept, set \( y = 0 \):
\[
2x + 1 = 0 \implies x = -\frac{1}{2}
\]
So, the x-intercept is \( (-\frac{1}{2}, 0) \).
- To find the y-intercept, set \( x = 0 \):
\[
-y + 1 = 0 \implies y = 1
\]
So, the y-intercept is \( (0, 1) \).
2. **For the line \( x + \lambda y - 3 = 0 \)**:
- To find the x-intercept, set \( y = 0 \):
\[
x - 3 = 0 \implies x = 3
\]
So, the x-intercept is \( (3, 0) \).
- To find the y-intercept, set \( x = 0 \):
\[
\lambda y - 3 = 0 \implies y = \frac{3}{\lambda}
\]
So, the y-intercept is \( (0, \frac{3}{\lambda}) \).
### Step 2: Identify the points of intersection.
The points of intersection of the lines with the axes are:
- From the first line: \( (-\frac{1}{2}, 0) \) and \( (0, 1) \)
- From the second line: \( (3, 0) \) and \( (0, \frac{3}{\lambda}) \)
### Step 3: Form the equation of the circle.
The circle that passes through these points can be represented as:
\[
(2x - y + 1)(x + \lambda y - 3) = 0
\]
### Step 4: Expand the equation.
Expanding the equation:
\[
(2x - y + 1)(x + \lambda y - 3) = 0
\]
Using distributive property:
\[
= 2x^2 + 2\lambda xy - 6x - xy - \lambda y^2 + 3y + x + \lambda y - 3 = 0
\]
Combining like terms:
\[
= 2x^2 + (2\lambda - 1)xy - \lambda y^2 - (6 - 3)x + (3 + \lambda)y - 3 = 0
\]
### Step 5: Apply the conditions for the circle.
For the equation to represent a circle, the coefficients of \( x^2 \) and \( y^2 \) must be equal, and the coefficient of \( xy \) must be zero:
1. Coefficient of \( x^2 \) is \( 2 \).
2. Coefficient of \( y^2 \) is \( -\lambda \).
3. Coefficient of \( xy \) is \( (2\lambda - 1) \).
Setting these conditions:
1. \( 2 = -\lambda \) implies \( \lambda = -2 \).
2. \( 2\lambda - 1 = 0 \) implies \( \lambda = \frac{1}{2} \).
### Conclusion
Thus, the values of \( \lambda \) are \( -2 \) and \( \frac{1}{2} \).
