Let a circle $$C$$ of radius 1 and closer to the origin be such that the lines passing through the point $$(3, 2)$$ and parallel to the coordinate axes touch it. Then the shortest distance of the circle $$C$$ from the point $$(5, 5)$$ is :

A circle of radius 1 is touched by lines through $$(3,2)$$ parallel to the coordinate axes; the one closest to the origin is sought. The lines through $$(3,2)$$ parallel to the axes are $$x = 3$$ and $$y = 2$$, and since the circle of radius 1 must touch both lines the center must be at distance 1 from each. Thus the possible centers are $$(3 \pm 1, 2 \pm 1) = (2,1), (2,3), (4,1), (4,3)$$.

To be closest to the origin the chosen center must minimize its distance from $$(0,0)$$:
- $$(2,1)$$: distance = $$\sqrt{5} \approx 2.24$$
- $$(2,3)$$: distance = $$\sqrt{13} \approx 3.61$$
- $$(4,1)$$: distance = $$\sqrt{17} \approx 4.12$$
- $$(4,3)$$: distance = $$5$$. The minimum occurs at $$(2,1)$$ as the center.

The shortest distance from this circle to the point $$(5,5)$$ is given by $$d = \text{distance from center to point} - r = \sqrt{(5-2)^2 + (5-1)^2} - 1 = \sqrt{9+16} - 1 = 5 - 1 = 4$$

The correct answer is Option (3): 4.