Let the locus of the mid points of the chords of circle $$x^2 + (y-1)^2 = 1$$ drawn from the origin intersect the line $$x + y = 1$$ at P and Q. Then, the length of PQ is:

The circle equation is $$x^2 + (y-1)^2 = 1$$, which expands to $$x^2 + y^2 - 2y = 0$$. The origin $$(0,0)$$ lies on this circle since $$0^2 + (0-1)^2 = 1$$.

Consider a chord drawn from the origin to another point $$(x_1, y_1)$$ on the circle. Let the midpoint of this chord be $$M(h, k)$$. Since $$M$$ is the midpoint, $$h = \frac{0 + x_1}{2} = \frac{x_1}{2}$$ and $$k = \frac{0 + y_1}{2} = \frac{y_1}{2}$$, so $$x_1 = 2h$$ and $$y_1 = 2k$$.

Substitute $$(x_1, y_1)$$ into the circle equation: $$(2h)^2 + (2k - 1)^2 = 1$$ $$4h^2 + 4k^2 - 4k + 1 = 1$$ $$4h^2 + 4k^2 - 4k = 0$$ Divide by 4: $$h^2 + k^2 - k = 0$$

Thus, the locus of the midpoints is $$x^2 + y^2 - y = 0$$. Completing the square: $$x^2 + y^2 - y = 0$$ $$x^2 + \left(y^2 - y + \frac{1}{4}\right) = \frac{1}{4}$$ $$x^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^2$$ This is a circle with center $$\left(0, \frac{1}{2}\right)$$ and radius $$\frac{1}{2}$$.

This locus circle intersects the line $$x + y = 1$$. Substitute $$y = 1 - x$$ into the locus equation: $$x^2 + \left((1 - x) - \frac{1}{2}\right)^2 = \frac{1}{4}$$ $$x^2 + \left(\frac{1}{2} - x\right)^2 = \frac{1}{4}$$ Expand $$\left(\frac{1}{2} - x\right)^2 = \frac{1}{4} - x + x^2$$: $$x^2 + \frac{1}{4} - x + x^2 = \frac{1}{4}$$ $$2x^2 - x + \frac{1}{4} = \frac{1}{4}$$ $$2x^2 - x = 0$$ $$x(2x - 1) = 0$$ So $$x = 0$$ or $$x = \frac{1}{2}$$.

Corresponding $$y$$-values: - If $$x = 0$$, $$y = 1 - 0 = 1$$, so point $$P(0, 1)$$. - If $$x = \frac{1}{2}$$, $$y = 1 - \frac{1}{2} = \frac{1}{2}$$, so point $$Q\left(\frac{1}{2}, \frac{1}{2}\right)$$.

The distance $$PQ$$ is: $$\sqrt{\left(\frac{1}{2} - 0\right)^2 + \left(\frac{1}{2} - 1\right)^2} = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$

Therefore, the length of $$PQ$$ is $$\frac{1}{\sqrt{2}}$$, corresponding to option A.