Consider a circle $$(x - \alpha)^2 + (y - \beta)^2 = 50$$, where $$\alpha, \beta > 0$$. If the circle touches the line $$y + x = 0$$ at the point P, whose distance from the origin is $$4\sqrt{2}$$, then $$(\alpha + \beta)^2$$ is equal to _____.

Circle $$(x-\alpha)^2 + (y-\beta)^2 = 50$$ with $$\alpha, \beta > 0$$ touches $$y + x = 0$$.

Distance from center $$(\alpha, \beta)$$ to the line $$x + y = 0$$:

$$\frac{|\alpha + \beta|}{\sqrt{2}} = \sqrt{50} = 5\sqrt{2}$$

Since $$\alpha, \beta > 0$$: $$\alpha + \beta = 10$$.

Point of tangency P has distance $$4\sqrt{2}$$ from origin.

P lies on $$x + y = 0$$, so $$P = (t, -t)$$ with $$|P| = |t|\sqrt{2} = 4\sqrt{2}$$, so $$|t| = 4$$, $$P = (4, -4)$$ or $$(-4, 4)$$.

P must be the foot of perpendicular from center to the line. The line from $$(\alpha, \beta)$$ perpendicular to $$x + y = 0$$ has direction $$(1, 1)$$.

$$P = (\alpha, \beta) - \frac{\alpha+\beta}{\sqrt{2}} \cdot \frac{(1,1)}{\sqrt{2}} = (\alpha - 5, \beta - 5)$$

If $$P = (4, -4)$$: $$\alpha = 9, \beta = 1$$. $$\alpha + \beta = 10$$ ✓.

If $$P = (-4, 4)$$: $$\alpha = 1, \beta = 9$$. $$\alpha + \beta = 10$$ ✓.

$$(\alpha + \beta)^2 = 100$$.

The answer is $$\boxed{100}$$.