If the circle `x^(2)+y^(2)+2ax=0, a in R` touches the parabola `y^(2)=4x`, them

Options

A. `a in (-oo, 0)`
B. `a in (0, oo)`
C. `a in (2, oo)`
D. none of these

Explanation

To solve the problem, we need to determine the conditions under which the circle given by the equation \(x^2 + y^2 + 2ax = 0\) touches the parabola given by the equation \(y^2 = 4x\).

### Step-by-Step Solution:

1. **Rewrite the Circle's Equation**:
The equation of the circle can be rewritten as:
\[ x^2 + y^2 = -2ax \]
This represents a circle centered at \((-a, 0)\) with radius \(\sqrt{a^2}\) (since the radius is derived from the standard form of a circle).

2. **Identify the Center and Radius**:
The center of the circle is at \((-a, 0)\) and the radius is \(|a|\). For the circle to touch the parabola, the center must lie on the negative x-axis, meaning \(a\) must be positive.

3. **Substituting the Parabola's Equation**:
The equation of the parabola is \(y^2 = 4x\). To find the points of intersection, we can substitute \(y^2\) from the parabola into the circle's equation.

From the parabola, we have:
\[ x = \frac{y^2}{4} \]
Substitute this into the circle's equation:
\[ \left(\frac{y^2}{4}\right)^2 + y^2 + 2a\left(\frac{y^2}{4}\right) = 0 \]

4. **Simplifying the Equation**:
Expanding the equation:
\[ \frac{y^4}{16} + y^2 + \frac{2ay^2}{4} = 0 \]
\[ \frac{y^4}{16} + y^2 + \frac{ay^2}{2} = 0 \]
Multiply through by 16 to eliminate the fraction:
\[ y^4 + 16y^2 + 8ay^2 = 0 \]
Combine like terms:
\[ y^4 + (16 + 8a)y^2 = 0 \]

5. **Factoring the Equation**:
Factor out \(y^2\):
\[ y^2(y^2 + (16 + 8a)) = 0 \]
This gives us two cases:
- \(y^2 = 0\) (which corresponds to the point of tangency)
- \(y^2 + (16 + 8a) = 0\) (this must have no real solutions for the circle to touch the parabola)

6. **Setting the Discriminant**:
For the quadratic \(y^2 + (16 + 8a) = 0\) to have no real solutions, the term \(16 + 8a\) must be less than or equal to zero:
\[ 16 + 8a \leq 0 \]
Solving this inequality:
\[ 8a \leq -16 \implies a \leq -2 \]

7. **Conclusion**:
Since \(a\) must also be positive for the center to lie on the negative x-axis, we conclude that:
\[ a > 0 \text{ and } a \leq -2 \]
This leads us to the conclusion that \(a\) must be in the range:
\[ a > 0 \]

### Final Answer:
The condition for the circle to touch the parabola is that \(a > 0\).