If there exists at least one point on the circle `x^(2)+y^(2)=a^(2)` from which two perpendicular tangents can be drawn to parabola `y^(2)=2x`, then find the values of a.

Explanation

To solve the problem, we need to determine the values of \( a \) such that there exists at least one point on the circle \( x^2 + y^2 = a^2 \) from which two perpendicular tangents can be drawn to the parabola \( y^2 = 2x \).

### Step 1: Identify the Directrix of the Parabola
The equation of the parabola is given by \( y^2 = 2x \). For a parabola of the form \( y^2 = 4ax \), the directrix is given by \( x = -a \). Here, we can identify \( 4a = 2 \), which gives us \( a = \frac{1}{2} \). Therefore, the directrix of our parabola is:
\[ x = -\frac{1}{2} \]

### Step 2: Equation of the Circle
The equation of the circle is given by:
\[ x^2 + y^2 = a^2 \]
This circle has a radius of \( a \).

### Step 3: Intersection of the Directrix and the Circle
To find the points of intersection between the directrix \( x = -\frac{1}{2} \) and the circle, we substitute \( x = -\frac{1}{2} \) into the circle's equation:
\[ \left(-\frac{1}{2}\right)^2 + y^2 = a^2 \]
This simplifies to:
\[ \frac{1}{4} + y^2 = a^2 \]
Rearranging gives:
\[ y^2 = a^2 - \frac{1}{4} \]

### Step 4: Condition for Real Roots
For the equation \( y^2 = a^2 - \frac{1}{4} \) to have real solutions, the right-hand side must be non-negative:
\[ a^2 - \frac{1}{4} \geq 0 \]
This leads to:
\[ a^2 \geq \frac{1}{4} \]

### Step 5: Taking the Square Root
Taking the square root of both sides, we find:
\[ |a| \geq \frac{1}{2} \]
This implies:
\[ a \geq \frac{1}{2} \quad \text{or} \quad a \leq -\frac{1}{2} \]

### Conclusion
Thus, the values of \( a \) for which there exists at least one point on the circle from which two perpendicular tangents can be drawn to the parabola are:
\[ a \geq \frac{1}{2} \quad \text{or} \quad a \leq -\frac{1}{2} \]