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

Options

A. `lamdagt0`
B. `lamdalt0`
C. `lamdagt1`
D. none of these

Explanation

To solve the problem of determining the conditions under which the circle \( x^2 + y^2 + 2\lambda x = 0 \) touches the parabola \( y^2 = 4x \) externally, we will follow these steps:

### Step 1: Rewrite the Circle Equation
The given equation of the circle is:
\[ x^2 + y^2 + 2\lambda x = 0 \]
We can rewrite it in standard form:
\[ x^2 + 2\lambda x + y^2 = 0 \]
This can be rearranged to:
\[ (x + \lambda)^2 + y^2 = \lambda^2 \]
This shows that the center of the circle is at \( (-\lambda, 0) \) and the radius is \( \lambda \).

### Step 2: Identify the Parabola
The equation of the parabola is:
\[ y^2 = 4x \]
This parabola opens to the right and has its vertex at the origin \( (0, 0) \).

### Step 3: Determine the 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 \( (0, 0) \).

The distance from the center of the circle \( (-\lambda, 0) \) to the origin \( (0, 0) \) is:
\[ \text{Distance} = \sqrt{(-\lambda - 0)^2 + (0 - 0)^2} = \sqrt{\lambda^2} = |\lambda| \]
Since the radius of the circle is \( \lambda \), for the circle to touch the parabola externally, we must have:
\[ |\lambda| = \lambda \]
This implies that \( \lambda \) must be greater than or equal to 0.

### Step 4: Ensure the Circle is on the Negative X-axis
For the circle to touch the parabola externally, the center of the circle must lie on the negative x-axis. This means:
\[ -\lambda < 0 \implies \lambda > 0 \]

### Conclusion
Thus, the condition for the circle to touch the parabola externally is:
\[ \lambda > 0 \]