Let $$y = mx + c$$, $$m > 0$$ be the focal chord of $$y^2 = -64x$$, which is tangent to $$(x+10)^2 + y^2 = 4$$. Then, the value of $$4\sqrt{2}(m+c)$$ is equal to ___.

The parabola $$y^2 = -64x$$ has $$4a = -64$$, so $$a = -16$$ and the focus is at $$(-16, 0)$$.

A focal chord passes through the focus, so substituting $$(-16, 0)$$ into $$y = mx + c$$ gives $$0 = -16m + c$$, hence $$c = 16m$$.

For the line $$y = mx + c$$ (i.e., $$mx - y + c = 0$$) to be tangent to the circle $$(x + 10)^2 + y^2 = 4$$ with centre $$(-10, 0)$$ and radius 2, the distance from the centre to the line must equal 2:

$$\frac{|m(-10) + c|}{\sqrt{m^2 + 1}} = 2$$

Substituting $$c = 16m$$: $$\frac{|-10m + 16m|}{\sqrt{m^2+1}} = \frac{6|m|}{\sqrt{m^2+1}} = 2$$

Squaring: $$36m^2 = 4(m^2 + 1) \implies 32m^2 = 4 \implies m^2 = \frac{1}{8} \implies m = \frac{1}{2\sqrt{2}}$$ (since $$m > 0$$).

Then $$c = 16m = \frac{16}{2\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}$$.

$$4\sqrt{2}(m + c) = 4\sqrt{2}\left(\frac{1}{2\sqrt{2}} + 4\sqrt{2}\right) = 4\sqrt{2} \cdot \frac{1}{2\sqrt{2}} + 4\sqrt{2} \cdot 4\sqrt{2} = 2 + 32 = 34$$