If the line y = mx + 4 is tangent to `x^(2) + y^(2) = 4` and `y^(2) = 4ax` then a is (a `gt` 0) equal to :

Options

A. `4 sqrt(3)`
B. `3 sqrt(2)`
C. `2 sqrt(2)`
D. `3 sqrt(3)`

Explanation

To solve the problem, we need to find the value of \( a \) given that the line \( y = mx + 4 \) is tangent to the circle \( x^2 + y^2 = 4 \) and the parabola \( y^2 = 4ax \).

### Step-by-Step Solution:

1. **Identify the Circle and its Radius**:
The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]
This can be rewritten in standard form as:
\[ x^2 + y^2 = r^2 \]
Here, \( r^2 = 4 \), thus the radius \( r \) is:
\[ r = 2 \]

2. **Condition for Tangency with the Circle**:
For the line \( y = mx + c \) to be tangent to the circle, the condition is:
\[ c = r \sqrt{1 + m^2} \]
Here, \( c = 4 \) (from the line equation) and \( r = 2 \). Therefore, we can set up the equation:
\[ 4 = 2 \sqrt{1 + m^2} \]

3. **Solve for \( m \)**:
Dividing both sides by 2:
\[ 2 = \sqrt{1 + m^2} \]
Now squaring both sides:
\[ 4 = 1 + m^2 \]
Rearranging gives:
\[ m^2 = 3 \]
Thus, we find:
\[ m = \pm \sqrt{3} \]

4. **Condition for Tangency with the Parabola**:
The equation of the parabola is:
\[ y^2 = 4ax \]
For the line \( y = mx + 4 \) to be tangent to the parabola, the condition is:
\[ c = \frac{a}{m} \]
Substituting \( c = 4 \):
\[ 4 = \frac{a}{m} \]
Rearranging gives:
\[ a = 4m \]

5. **Substituting Values of \( m \)**:
We have two possible values for \( m \):
- \( m = \sqrt{3} \)
- \( m = -\sqrt{3} \)

For \( m = \sqrt{3} \):
\[ a = 4(\sqrt{3}) = 4\sqrt{3} \]

For \( m = -\sqrt{3} \):
\[ a = 4(-\sqrt{3}) = -4\sqrt{3} \]

6. **Final Condition**:
Since the problem states that \( a > 0 \), we discard \( -4\sqrt{3} \) and conclude:
\[ a = 4\sqrt{3} \]

### Conclusion:
The value of \( a \) is:
\[ \boxed{4\sqrt{3}} \]