I know from visual inspection that to find the shortest distance between two non-intersecting circles as shown in diagram below, one needs to connect their centers and then find the distance between the points where this segment intersects the two circles.
But I could not come up with a geometrical proof for this fact.
Question
How would I go about proving the above fact? Any hint would be helpful.
Best Answer
Assume the shortest distance is attained between two other points $C', D'$ on the respective circles, so that $C'D' \lt CD$. Since $AC=AC'$ and $BD=BD'$ that would imply:
$$ AC'+C'D'+D'B \lt AC'+CD+D'B = AC+CD+DB=AB $$
But the shortest distance between two points is the straight line, so the length of the broken line $AC'D'B$ can be no smaller than that of segment $AB\,$, with equality iff $C'=D$, $D'=D$.