The circle `x^(2)+y^(2)+2lamdax=0,lamdainR` touches the parabola `y^(2)=4x` externally. Then

Options

A. `lamda=-1`
B. `lamda=1`
C. `lamda=2`
D. `lamdalt-1`

Answer

To solve the problem, we need to find the values of \( \lambda \) such that the circle defined by the equation \( x^2 + y^2 + 2\lambda x = 0 \) touches the parabola defined by \( y^2 = 4x \) externally.

### Step-by-Step Solution:

1. **Rewrite the Circle's Equation**:
The given equation of the circle is:
\[ x^2 + y^2 + 2\lambda x = 0 \]
We can rearrange this to:
\[ x^2 + 2\lambda x + y^2 = 0 \]
To make it more recognizable, we complete the square for the \( x \) terms:
\[ (x + \lambda)^2 + y^2 = \lambda^2 \]
This shows that the circle has:
- Center: \( (-\lambda, 0) \)
- Radius: \( |\lambda| \)

2. **Identify the Parabola**:
The equation of the parabola is:
\[ y^2 = 4x \]
This parabola opens to the right with its vertex at the origin.

3. **Condition for External Tangency**:
For the circle to touch the parabola externally, the distance from the center of the circle to the vertex of the parabola must equal the radius of the circle. The vertex of the parabola is at the origin \( (0, 0) \).

4. **Calculate the Distance**:
The distance \( d \) from the center of the circle \( (-\lambda, 0) \) to the vertex \( (0, 0) \) is given by:
\[ d = \sqrt{(-\lambda - 0)^2 + (0 - 0)^2} = |\lambda| \]

5. **Set Up the Equation**:
For the circle to touch the parabola externally, we set the distance equal to the radius:
\[ |\lambda| = |\lambda| \]
This condition is always satisfied, but we need to ensure that the center of the circle is positioned correctly.

6. **Determine the Sign of \( \lambda \)**:
Since the circle must touch the parabola externally, the center of the circle must be on the negative side of the x-axis. Thus, we require:
\[ -\lambda > 0 \quad \Rightarrow \quad \lambda < 0 \]

7. **Conclusion**:
The values of \( \lambda \) that satisfy the condition for external tangency are:
\[ \lambda < 0 \]
Therefore, \( \lambda \) can take any negative value.