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. `4sqrt3`
B. `3sqrt2`
C. `2sqrt2`
D. `3sqrt3`

Explanation

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

### Step-by-Step Solution:

1. **Identify the Circle's Properties**:
The equation of the circle is \( x^2 + y^2 = 4 \).
- The radius \( R \) of the circle is \( \sqrt{4} = 2 \).
- The y-intercept \( C \) of the line \( y = mx + 4 \) is \( 4 \).

2. **Use the Tangent Condition for the Circle**:
The condition for a line to be tangent to a circle is given by:
\[ C = R \sqrt{1 + m^2} \]
Substituting the known values:
\[ 4 = 2 \sqrt{1 + m^2} \]

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

4. **Use the Tangent Condition for the Parabola**:
The equation of the parabola is \( y^2 = 4ax \). The condition for a line to be tangent to a parabola is:
\[ C = \frac{a}{m} \]
Substituting \( C = 4 \) and \( m = \pm \sqrt{3} \):
\[ 4 = \frac{a}{\sqrt{3}} \]

5. **Solve for \( a \)**:
Rearranging gives:
\[ a = 4\sqrt{3} \]

6. **Conclusion**:
Since \( a \) must be greater than 0, we take the positive value:
\[ a = 4\sqrt{3} \]

### Final Answer:
Thus, the value of \( a \) is \( 4\sqrt{3} \).