Let L$$_1$$ be the length of the common chord of the curves $$x^2 + y^2 = 9$$ and $$y^2 = 8x$$, and L$$_2$$ be the length of the latus rectum of $$y^2 = 8x$$, then:

We have two curves. The first is the circle $$x^2 + y^2 = 9$$ whose centre is at the origin and whose radius is $$r = 3$$ because, for a circle $$x^2 + y^2 = R^2$$, the radius is $$R$$. The second curve is the parabola $$y^2 = 8x$$; this is of the standard form $$y^2 = 4ax$$ where $$a$$ is the focal length.

First we find the length $$L_2$$ of the latus rectum of the parabola. For any parabola of the form $$y^2 = 4ax$$, the length of the latus rectum is given by the formula $$4a$$. Comparing $$y^2 = 8x$$ with $$y^2 = 4ax$$ we see

$$4a = 8 \; \Longrightarrow \; a = 2.$$

Therefore the length of the latus rectum is

$$L_2 = 4a = 4 \times 2 = 8.$$

Now we calculate $$L_1$$, the length of the common chord of the circle and the parabola. Points on both curves must satisfy both equations simultaneously. From the parabola we have $$y^2 = 8x$$, so we substitute $$y^2$$ from this relation into the circle’s equation.

Circle equation: $$x^2 + y^2 = 9.$$

Substituting $$y^2 = 8x$$ gives

$$x^2 + 8x = 9.$$

Bringing all terms to the left side we obtain a quadratic in $$x$$,

$$x^2 + 8x - 9 = 0.$$

To solve this quadratic we use the quadratic formula. For $$ax^2 + bx + c = 0$$, the roots are

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$

Here $$a = 1, \; b = 8, \; c = -9$$, so

$$x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} = \frac{-8 \pm \sqrt{64 + 36}}{2} = \frac{-8 \pm \sqrt{100}}{2} = \frac{-8 \pm 10}{2}.$$

This gives two algebraic values

$$x_1 = \frac{-8 + 10}{2} = \frac{2}{2} = 1, \quad x_2 = \frac{-8 - 10}{2} = \frac{-18}{2} = -9.$$

However, for the parabola $$y^2 = 8x$$ we must have $$x \ge 0$$ because $$y^2$$ is non-negative. Thus $$x = -9$$ is inadmissible (it would force $$y^2 = -72$$, which is impossible for real $$y$$). We therefore keep only

$$x = 1.$$

For this $$x$$-coordinate, the corresponding $$y$$-coordinates come from $$y^2 = 8x$$:

$$y^2 = 8(1) = 8 \; \Longrightarrow \; y = \pm \sqrt{8} = \pm 2\sqrt{2}.$$

Hence the two real points of intersection are

$$(1,\, 2\sqrt{2}) \quad \text{and} \quad (1,\,-2\sqrt{2}).$$

The common chord is simply the straight line segment joining these two points. Because their $$x$$-coordinates are identical (both equal to 1), the chord is vertical. The distance between two points $$(x, y_1)$$ and $$(x, y_2)$$ on a vertical line is $$|y_1 - y_2|.$$ Therefore the length of the chord is

$$L_1 = |\,2\sqrt{2} - (-2\sqrt{2})| = |\,4\sqrt{2}\,| = 4\sqrt{2}.$$

We now compare $$L_1$$ and $$L_2$$:

$$L_1 = 4\sqrt{2}, \qquad L_2 = 8.$$

Dividing gives

$$\frac{L_1}{L_2} = \frac{4\sqrt{2}}{8} = \frac{\sqrt{2}}{2} \approx 0.707,$$

which is clearly less than 1. That is, $$L_1 < L_2.$$

Hence, the correct answer is Option C.