If 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 then the value of `lambda ` is `:`

Options

A. `1//2`
B. 2
C. 1
D. `-2`

Answer

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.

### Step 1: Find the points of intersection of the lines with the axes.

1. **For the first 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 \( A(-\frac{1}{2}, 0) \).
- To find the y-intercept, set \( x = 0 \):
\[ -y + 1 = 0 \implies y = 1 \]
So, the y-intercept is \( B(0, 1) \).

2. **For the second 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 \( C(3, 0) \).
- To find the y-intercept, set \( x = 0 \):
\[ \lambda y - 3 = 0 \implies y = \frac{3}{\lambda} \]
So, the y-intercept is \( D(0, \frac{3}{\lambda}) \).

### Step 2: Use the property of concyclic points.

The points \( A \), \( B \), \( C \), and \( D \) are concyclic, meaning that the product of the lengths of the segments from the center of the circle to these points must satisfy the condition:
\[ OA \cdot OC = OB \cdot OD \]

Where:
- \( OA = \frac{1}{2} \) (distance from origin to A)
- \( OB = 1 \) (distance from origin to B)
- \( OC = 3 \) (distance from origin to C)
- \( OD = \frac{3}{\lambda} \) (distance from origin to D)

### Step 3: Set up the equation.

Using the distances:
\[ \left(\frac{1}{2}\right) \cdot 3 = 1 \cdot \left(\frac{3}{\lambda}\right) \]

### Step 4: Solve for \( \lambda \).

1. Simplifying the left side:
\[ \frac{3}{2} = \frac{3}{\lambda} \]
2. Cross-multiplying gives:
\[ 3\lambda = 6 \implies \lambda = 2 \]

### Conclusion

Thus, the value of \( \lambda \) is \( 2 \).

---