A line is a common tangent to the circle `(x-3)^2+y^2=9` and the parabola `y^2=4x`. If the two points of contact (a,b) and (c,d) are distinct and lie in the first quadrant , then `2(a+c)` is equal to _______ .

Answer

To solve the problem, we need to find the value of \(2(a+c)\) where \((a,b)\) and \((c,d)\) are the points of tangency of a common tangent line to the given circle and parabola.

### Step 1: Identify the Circle and Parabola
The equation of the circle is given by:
\[ (x-3)^2 + y^2 = 9 \]
This can be rewritten as:
\[ (x-3)^2 + y^2 = 3^2 \]
From this, we can identify that the center of the circle is at \((3, 0)\) and the radius \(r = 3\).

The equation of the parabola is given by:
\[ y^2 = 4x \]
This is a standard parabola that opens to the right with vertex at the origin \((0, 0)\).

### Step 2: Parametric Form of the Parabola
The parametric equations for points on the parabola \(y^2 = 4x\) can be expressed as:
\[ (x, y) = (t^2, 2t) \]
where \(t\) is a parameter.

### Step 3: Equation of the Tangent Line
The equation of the tangent line to the parabola at the point \((t^2, 2t)\) is given by:
\[ ty = x + t^2 \]
Rearranging gives:
\[ x - ty + t^2 = 0 \]

### Step 4: Perpendicular Distance from Circle Center to the Tangent Line
The distance \(d\) from the center of the circle \((3, 0)\) to the tangent line \(x - ty + t^2 = 0\) must equal the radius \(3\). The formula for the distance from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line, \(A = 1\), \(B = -t\), and \(C = t^2\). Thus, the distance becomes:
\[ d = \frac{|3 - 0 + t^2|}{\sqrt{1 + t^2}} = \frac{|3 + t^2|}{\sqrt{1 + t^2}} \]
Setting this equal to the radius:
\[ \frac{|3 + t^2|}{\sqrt{1 + t^2}} = 3 \]

### Step 5: Solve for \(t\)
Squaring both sides gives:
\[ (3 + t^2)^2 = 9(1 + t^2) \]
Expanding both sides:
\[ 9 + 6t^2 + t^4 = 9 + 9t^2 \]
Rearranging leads to:
\[ t^4 - 3t^2 = 0 \]
Factoring out \(t^2\):
\[ t^2(t^2 - 3) = 0 \]
This gives us:
\[ t^2 = 0 \quad \text{or} \quad t^2 = 3 \]
Since \(t^2 = 0\) corresponds to the point \((0, 0)\) which is not in the first quadrant, we take \(t^2 = 3\).

### Step 6: Calculate Points of Tangency
For \(t = \sqrt{3}\):
\[ a = t^2 = 3 \quad \text{and} \quad b = 2t = 2\sqrt{3} \]
The points of tangency are:
\[ (a, b) = (3, 2\sqrt{3}) \]

For \(t = -\sqrt{3}\):
\[ c = t^2 = 3 \quad \text{and} \quad d = 2t = -2\sqrt{3} \]
Since we need points in the first quadrant, we discard this point.

### Step 7: Calculate \(2(a+c)\)
Since both points of tangency have the same \(x\)-coordinate:
\[ 2(a+c) = 2(3 + 3) = 2 \times 6 = 12 \]

### Final Answer
Thus, the value of \(2(a+c)\) is:
\[ \boxed{12} \]