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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.

Shortest distance between two points on 2 non-intersecting circles]

Sunil
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  • See whether this helps you: http://math.stackexchange.com/questions/1688605/shortest-distance-between-two-circles –  Feb 01 '17 at 09:13
  • Thanks. I will have a look at the explanation by Vincent in that post. I was thinking that this is a standard theorem in Geometry, but may be not. – Sunil Feb 01 '17 at 09:16
  • if a curve is differentiable, and P is the closest point on that curve to the point X, then the line tangent to the curve at the point P will be perpendicular to the line segment XP. – Steven Alexis Gregory Jul 20 '19 at 21:28

3 Answers3

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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 $C'=D$, $D'=D$.

dxiv
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Not phrasing this as a formal proof, but if you draw the tangent lines at $C$ and at $D$ the circles will lie entirely in opposite halfplanes.

Thus, any other segment joining a point on one circle and a point on the other will be longer than the distance between $C$ and $D$ (because tangent lines are perpendicular to radii) which realizes the minimum.

AdLibitum
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If there are two points $(1,2) $ on two curves symmetric to $y-$ axis as $ y_1(x), y_2(x) $ satisfying max/min derivative conditions:

$$ x_1=0,y_1^\prime =0 ,\, y_1^{\prime\prime } >0 ;x_2=0,\, y_2^\prime = 0,y_1^{\prime\prime }< 0 , $$

then the minimum distance between points $1,2$ is $|y_1- y_1|$ occurring along line of symmetry.. the $y-$ axis.

This holds good for such arbitrary curves and rotations in the $xy$ plane resulting in particular the example you gave.

CircTgts

At these tangent points direction of normal is same and tangents are parallel.

Narasimham
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