If a circle C, whose radius is 3, touches externally the circle $$x^2 + y^2 + 2x - 4y - 4 = 0$$ at the point (2, 2), then the length of the intercept cut by this circle C on the x-axis is equal to:

We begin with the given circle $$x^2 + y^2 + 2x - 4y - 4 = 0.$$

For any circle written in the general form $$x^2 + y^2 + 2gx + 2fy + c = 0,$$ the centre is $$(-g,\,-f)$$ and the radius is $$r=\sqrt{g^2 + f^2 - c}.$$

Comparing, we have $$2g = 2 \;\Longrightarrow\; g = 1,$$ and $$2f = -4 \;\Longrightarrow\; f = -2,$$ while $$c = -4.$$

So the centre of the given circle is $$(-g,\,-f)=(-1,\,2)$$ and its radius is

$$r_1 \;=\;\sqrt{g^2 + f^2 - c} \;=\; \sqrt{1^2 + (-2)^2 - (-4)} \;=\; \sqrt{1 + 4 + 4} \;=\; \sqrt{9} \;=\; 3.$$

The new circle $$C$$ also has radius $$3$$ and touches the given circle externally at the point $$(2,\,2).$$

When two equal circles touch externally, the point of contact is exactly the midpoint of the line segment joining their centres.

Let the centre of circle $$C$$ be $$(h,\,k).$$ Because $$(2,\,2)$$ is the midpoint of $$(-1,\,2)$$ and $$(h,\,k),$$ we write the midpoint relations:

$$\dfrac{-1 + h}{2} = 2,\qquad \dfrac{\,2 + k\,}{2} = 2.$$

Solving,

$$-1 + h = 4 \;\Longrightarrow\; h = 5,$$

$$2 + k = 4 \;\Longrightarrow\; k = 2.$$

Thus the centre of circle $$C$$ is $$(5,\,2),$$ and its equation is

$$ (x - 5)^2 \;+\; (y - 2)^2 \;=\; 3^2 \;=\; 9. $$

To find the intercept cut by this circle on the x-axis, we set $$y = 0$$ (because every point on the x-axis has zero y-coordinate) and solve for $$x.$$

Substituting $$y = 0$$ in the equation,

$$ (x - 5)^2 + (0 - 2)^2 = 9, $$

so

$$ (x - 5)^2 + 4 = 9 \;\Longrightarrow\; (x - 5)^2 = 5. $$

Taking square roots,

$$ x - 5 = \pm\sqrt{5} \;\Longrightarrow\; x = 5 \pm \sqrt{5}. $$

Hence the points of intersection with the x-axis are $$(5 - \sqrt{5},\,0)$$ and $$(5 + \sqrt{5},\,0).$$

The length of the intercept is the distance between these two x-coordinates:

$$ (5 + \sqrt{5}) - (5 - \sqrt{5}) = 2\sqrt{5}. $$

Hence, the correct answer is Option D.