A triangle is formed by the tangents at the point $$(2, 2)$$ on the curves $$y^2 = 2x$$ and $$x^2 + y^2 = 4x$$, and the line $$x + y + 2 = 0$$. If $$r$$ is the radius of its circumcircle, then $$r^2$$ is equal to

We need to find the circumradius squared of the triangle formed by three lines.

First, consider the tangent to $$y^2 = 2x$$ at $$(2,2)$$. Differentiating $$y^2 = 2x$$ gives $$2y \frac{dy}{dx} = 2$$, so $$\frac{dy}{dx} = \frac{1}{y} = \frac{1}{2}$$ at $$(2,2)$$. Therefore, the tangent line is $$y - 2 = \frac{1}{2}(x - 2)$$, which can be written as $$x - 2y + 2 = 0$$.

Next, consider the tangent to $$x^2 + y^2 = 4x$$ at $$(2,2)$$. Rewriting the equation as $$(x-2)^2 + y^2 = 4$$ shows that this circle has center $$(2,0)$$ and radius 2. The radius drawn to $$(2,2)$$ is vertical, which means the tangent there is horizontal: $$y = 2$$.

The third line is given by $$x + y + 2 = 0$$.

The intersections of these three lines determine the vertices of the triangle. Solving $$x - 2y + 2 = 0$$ with $$y = 2$$ yields $$x = 2$$, so vertex $$A$$ is $$(2,2)$$. Solving $$x - 2y + 2 = 0$$ with $$x + y + 2 = 0$$ (i.e., $$x = -y - 2$$) gives $$y = 0$$ and $$x = -2$$, so vertex $$B$$ is $$(-2,0)$$. Finally, solving $$y = 2$$ with $$x + y + 2 = 0$$ gives $$x = -4$$, so vertex $$C$$ is $$(-4,2)$$.

Let $$a = BC$$, $$b = AC$$, and $$c = AB$$. Then

$$a^2 = BC^2 = (-2+4)^2 + (0-2)^2 = 4 + 4 = 8$$

$$b^2 = AC^2 = (2+4)^2 + (2-2)^2 = 36$$

$$c^2 = AB^2 = (2+2)^2 + (2-0)^2 = 16 + 4 = 20$$

The area of the triangle can be found by the determinant formula: $$\text{Area} = \frac{1}{2}|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$$, which becomes $$\frac{1}{2}|2(0-2) + (-2)(2-2) + (-4)(2-0)| = 6$$.

Finally, the circumradius is given by $$R = \frac{abc}{4 \cdot \text{Area}}$$, so

$$R^2 = \frac{a^2 b^2 c^2}{16 \cdot \text{Area}^2} = \frac{8 \times 36 \times 20}{16 \times 36} = \frac{5760}{576} = 10$$

Therefore, $$r^2 = \boxed{10}$$.