Equation of a common tangent to the parabola $$y^2 = 4x$$ and the hyperbola $$xy = 2$$ is:

First recall the standard facts about the given curves.

The parabola is $$y^{2}=4x$$. Comparing this with the general form $$y^{2}=4ax$$, we see that the parameter is $$a=1$$.

For a parabola $$y^{2}=4ax$$, the equation of a tangent having slope $$m$$ is given by the slope-form formula

$$y=mx+\dfrac{a}{m}.$$

Stating the value of $$a$$ and substituting it, we get

$$y = mx + \dfrac{1}{m} \quad\text{(1)}$$

This straight line touches the parabola by construction. We now require the same line to be a tangent to the hyperbola $$xy = 2$$ as well.

To impose this condition, we substitute the expression for $$y$$ from equation (1) into the hyperbola’s equation:

$$x\left(mx+\dfrac{1}{m}\right)=2.$$

Multiplying out, we obtain a quadratic in $$x$$:

$$m x^{2} + \dfrac{1}{m}\,x - 2 = 0.$$

For the line to be tangent to the hyperbola, this quadratic must possess exactly one real root. The condition for a quadratic $$Ax^{2}+Bx+C=0$$ to have equal (coincident) roots is that its discriminant vanishes, that is

$$B^{2}-4AC = 0.$$

Here $$A = m,\; B = \dfrac{1}{m},\; C = -2.$$ Substituting these values, we write the discriminant:

$$\left(\dfrac{1}{m}\right)^{2} - 4\,(m)\,(-2) = 0.$$

Simplifying term by term,

$$\dfrac{1}{m^{2}} + 8m = 0.$$

To clear the fraction, multiply the entire equation by $$m^{2}$$:

$$1 + 8m^{3} = 0.$$

Thus

$$8m^{3} = -1 \quad\Longrightarrow\quad m^{3} = -\dfrac{1}{8}.$$

Taking the real cube root on both sides, we find

$$m = -\dfrac{1}{2}.$$

Now we substitute this value of $$m$$ back into the tangent equation (1):

$$y = \left(-\dfrac{1}{2}\right)x + \dfrac{1}{\, -\dfrac{1}{2}\,}.$$

Calculating the constant term,

$$\dfrac{1}{\, -\dfrac{1}{2}\,} = -2,$$

so the equation becomes

$$y = -\dfrac{1}{2}x - 2.$$

To express this in the usual linear form, multiply through by $$2$$:

$$2y = -x - 4.$$

Bringing all terms to the left side gives

$$x + 2y + 4 = 0.$$

This matches Option C.

Hence, the correct answer is Option C.