Two tangent lines $$l_1$$ and $$l_2$$ are drawn from the point (2, 0) to the parabola $$2y^2 = -x$$. If the lines $$l_1$$ and $$l_2$$ are also tangent to the circle $$(x-5)^2 + y^2 = r$$, then $$17r^2$$ is equal to

We have the parabola $$2y^2 = -x$$, i.e., $$y^2 = -x/2$$. Comparing with $$y^2 = 4ax$$, we get $$a = -1/8$$.

The tangent in slope form is $$y = mx + \frac{a}{m} = mx - \frac{1}{8m}$$. Since it passes through $$(2, 0)$$:

$$0 = 2m - \frac{1}{8m} \implies 16m^2 = 1 \implies m = \pm\frac{1}{4}$$

So the two tangent lines are $$l_1: x - 4y - 2 = 0$$ and $$l_2: x + 4y - 2 = 0$$.

These are also tangent to the circle $$(x - 5)^2 + y^2 = r$$, where $$r$$ is the radius. The tangency condition requires the distance from centre $$(5, 0)$$ to each line to equal $$r$$:

$$r = \frac{|5 - 0 - 2|}{\sqrt{1 + 16}} = \frac{3}{\sqrt{17}}$$

Therefore $$r^2 = \frac{9}{17}$$, and $$17r^2 = 9$$.

Hence, the correct answer is 9.