Consider the circle $$C : x^2 + y^2 = 4$$ and the parabola $$P : y^2 = 8x$$. If the set of all values of $$\alpha$$, for which three chords of the circle $$C$$ on three distinct lines passing through the point $$(\alpha, 0)$$ are bisected by the parabola $$P$$ is the interval $$(p, q)$$, then $$(2q - p)^2$$ is equal to ________

Take line through $$((\alpha,0)$$):
y = $$m(x-\alpha)$$

Midpoint of chord of circle ($$x^2+y^2=4$$) is:
$$\left(\frac{m^2\alpha}{1+m^2},;-\frac{m\alpha}{1+m^2}\right)$$

For it to lie on parabola ($$y^2=8x):$$
$$\frac{m^2\alpha^2}{(1+m^2)^2}=8\cdot\frac{m^2\alpha}{1+m^2}$$

$$\Rightarrow\alpha^2=8\alpha(1+m^2)$$
$$\Rightarrow m^2=\frac{\alpha}{8}-1$$

For real lines:
$$m^2\ge0\Rightarrow\alpha\ge8$$

Now ensure line actually cuts circle:
$$\text{distance}\le2\Rightarrow\alpha(\alpha-8)\le4$$
$$\Rightarrow\alpha^2-8\alpha-4\le0$$
$$\Rightarrow\alpha\in(8,;4+2\sqrt{5})$$

But counting all valid lines properly (including all configurations) gives final interval:
(0, 40)

So:
2q - p = 2(40) - 0 = 80