If the circle `x^2 + y^2 +2lamdax=0,lamda``in` R touches the parabola `y^2 = 4x` externally, then

Options

A. `lamda=1`
B. `lamdagt1`
C. `lamdagt0`
D. `lamdalt0`

Answer

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

### Step 1: Rewrite the Circle Equation
The equation of the circle can be rewritten in standard form. The given equation is:
\[ x^2 + y^2 + 2\lambda x = 0 \]
This can be rearranged as:
\[ x^2 + 2\lambda x + y^2 = 0 \]
Completing the square for \(x\):
\[ (x + \lambda)^2 + y^2 = \lambda^2 \]
This shows that the circle has a center at \((- \lambda, 0)\) and a radius of \(|\lambda|\).

### Step 2: Identify the Parabola
The parabola given 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.

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

2. **Radius of the circle:**
The radius of the circle is \(|\lambda|\).

For the circle to touch the parabola externally, we need:
\[ d = \text{radius} \]
This gives us:
\[ |\lambda| = |\lambda| \]
This condition is always true, but we need to ensure that the circle is positioned correctly.

### Step 4: Analyze the Position of the Circle
Since the center of the circle is at \((- \lambda, 0)\), for the circle to touch the parabola externally, we need \(-\lambda\) to be less than 0 (which means \(\lambda > 0\)).

### Conclusion
Thus, the condition for \(\lambda\) is:
\[ \lambda > 0 \]