Let a circle $$C_1$$ be obtained on rolling the circle $$x^2 + y^2 - 4x - 6y + 11 = 0$$ upwards 4 units on the tangent T to it at the point (3, 2). Let $$C_2$$ be the image of $$C_1$$ in T. Let $$A$$ and $$B$$ be the centers of circles $$C_1$$ and $$C_2$$ respectively, and $$M$$ and $$N$$ be respectively the feet of perpendiculars drawn from $$A$$ and $$B$$ on the x-axis. Then the area of the trapezium AMNB is:

Find the area of trapezium AMNB where A, B are centers of circles $$C_1$$, $$C_2$$ and M, N are their feet of perpendiculars on the x-axis.

The given circle can be written as $$x^2 + y^2 - 4x - 6y + 11 = 0 \Rightarrow (x-2)^2 + (y-3)^2 = 2$$, which shows that its center is $$(2, 3)$$ and its radius is $$\sqrt{2}$$.

The tangent at the point $$(3, 2)$$ on this circle is obtained from the formula $$(x-2)(3-2) + (y-3)(2-3) = 2$$, which simplifies to $$(x-2) - (y-3) = 2$$ or $$x - y = 1$$.

When the circle rolls 4 units upward along this tangent line, its center moves in the direction of the unit vector along $$x - y = 1$$, namely $$\left(\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}}\right)$$. Hence the new center A of circle $$C_1$$ is $$A = \left(2 + \frac{4}{\sqrt{2}},\,3 + \frac{4}{\sqrt{2}}\right) = (2 + 2\sqrt{2},\,3 + 2\sqrt{2}).$$

The image of this center in the line $$x - y - 1 = 0$$ gives the center B of circle $$C_2$$. Reflecting $$(a, b)$$ in this line yields $$x' = b + 1,\quad y' = a - 1,$$ so $$B = (3 + 2\sqrt{2} + 1,\;2 + 2\sqrt{2} - 1) = (4 + 2\sqrt{2},\,1 + 2\sqrt{2}).$$

The feet of the perpendiculars from A and B onto the x-axis are $$M = (2 + 2\sqrt{2},\,0)$$ and $$N = (4 + 2\sqrt{2},\,0),$$ respectively.

The heights of A and B above the x-axis are $$AM = 3 + 2\sqrt{2},\quad BN = 1 + 2\sqrt{2},$$ and the distance between M and N is $$MN = (4 + 2\sqrt{2}) - (2 + 2\sqrt{2}) = 2.$$ Therefore the area of trapezium AMNB is $$\frac{1}{2}(AM + BN)\times MN = \frac{1}{2}(3 + 2\sqrt{2} + 1 + 2\sqrt{2})\times 2 = 4 + 4\sqrt{2} = 4(1 + \sqrt{2}).$$

The correct answer is Option B: $$4(1 + \sqrt{2})$$.