A circle of radius $$2$$ unit passes through the vertex and the focus of the parabola $$y^2 = 2x$$ and touches the parabola $$y = \left(x - \frac{1}{4}\right)^2 + \alpha$$, where $$\alpha > 0$$. Then $$(4\alpha - 8)^2$$ is equal to ______

A circle of radius 2 passes through the vertex and focus of the parabola $$y^2 = 2x$$ and touches the parabola $$y = \left(x - \frac{1}{4}\right)^2 + \alpha$$ where $$\alpha > 0$$. Find $$(4\alpha - 8)^2$$.

Comparing $$y^2 = 2x$$ with $$y^2 = 4ax$$, we find $$4a = 2$$, so $$a = \frac{1}{2}$$. Therefore the vertex is $$(0,0)$$ and the focus is $$\left(\frac{1}{2}, 0\right)$$.

Next, let the center of the desired circle be $$(h, k)$$. Since it passes through $$(0,0)$$ and $$\left(\frac{1}{2}, 0\right)$$ and has radius 2, we have the distance conditions:

$$h^2 + k^2 = 4 \quad \cdots (1)$$

and

$$ (h - \tfrac{1}{2})^2 + k^2 = 4 \quad \cdots (2)$$

Subtracting (2) from (1) yields $$h^2 - (h - \tfrac{1}{2})^2 = 0$$, which gives $$h - \tfrac{1}{4} = 0$$ and hence $$h = \frac{1}{4}$$. Substituting into (1) then yields $$k^2 = 4 - \frac{1}{16} = \frac{63}{16}$$, so $$k = \pm\frac{3\sqrt{7}}{4}$$.

Thus there are two possible centers, $$\left(\frac{1}{4}, \frac{3\sqrt{7}}{4}\right)$$ (above the $$x$$-axis) and $$\left(\frac{1}{4}, -\frac{3\sqrt{7}}{4}\right)$$ (below the $$x$$-axis).

Since the parabola $$y = \left(x - \frac{1}{4}\right)^2 + \alpha$$ opens upward with vertex at $$\left(\frac{1}{4}, \alpha\right)$$ and axis of symmetry $$x = \frac{1}{4}$$, and since both circle centers lie on this axis, any tangency must occur at $$x = \frac{1}{4}$$. At this value of $$x$$, the circle has $$y = k \pm 2$$, while the parabola has $$y = \alpha$$.

Because the parabola opens upward, the relevant circle must lie below it, so its topmost point touches the parabola. For the circle centered at $$\left(\frac{1}{4}, -\frac{3\sqrt{7}}{4}\right)$$, the topmost point is at

$$y = -\frac{3\sqrt{7}}{4} + 2 = \frac{8 - 3\sqrt{7}}{4}.$$

Since $$3\sqrt{7} \approx 7.937$$, this value is positive (about $$0.016$$), and hence we take $$\alpha = \frac{8 - 3\sqrt{7}}{4} \;>\; 0$$. Because both the circle and the parabola have horizontal tangents at this point and the circle curves downward while the parabola curves upward, they touch without crossing.

Finally, since $$\alpha = \frac{8 - 3\sqrt{7}}{4}$$, we have $$4\alpha = 8 - 3\sqrt{7}$$, so $$4\alpha - 8 = -3\sqrt{7}$$, and

$$ (4\alpha - 8)^2 = 9 \times 7 = 63. $$

The answer is $$\boxed{63}$$.