Going back and re-reading the original problem, there's no constraint on the direction of the "perfectly straight" path. The path could be tangential to the 1km circle or it could point radially outward. Given that, here's the worst case. Imagine you walk due north to the 1km circle, stop, turn 90 degrees left, and start tracing out the circle. Imagine the path actually started just to your right and had a heading toward the left, just a few degrees above being tangential with the circle. In this case, any divergence you make from the circle until you get all the way around will guarantee that you never interesect with the path to safety.

If you're trying to minimize your worst-case walking distance, then the simple answer (walk the full circle) seems to be the right answer. However, if we change the question around to be minimizing your *expected* case walking distance, given an equal probability of the safe road being at any starting point in any direction, then things get more interesting. The concern, if you stick on the 1km circle, is that you don't get to "cut off the corner" by intersecting with the safe path somewhere beyond the 1km radius.

I suspect that a useful strategy would be some kind of increasing radius arc. You need to solve for d radius / d theta, such that the expected length of the walk is minimized. To compute the expectation, you've got to integrate over all possible paths. You also have to know how long the path is that takes you to safety, such that you don't overshoot wherever that safe path takes you. I wonder, however, whether you'd just end up with d radius / d theta = 0, getting you back to the degenerate, simple solution. (Also, since the length of the safe path isn't specified, I'll bet the problem's author never considered this sort of solution.)