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Suppose we have two circles with radii $\delta>0$ and the distance between circle is $r$, i.e. $|AB|=r$. Let $x$ and $y$ be points in distinct circles. How to prove rigorously that $|x-y|>r$?

I've tried to use triangle inequality but it did not work out. I believe that it should be simple but it is not clear to me.

Thank you!

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RFZ
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    @TedShifrin, I think you meant that $x$ and $y$ are on the line connecting $A$ and $B$ but not on the line segment $AB$ since $x,y$ are inside circles. – RFZ Aug 16 '22 at 23:34
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    I didn’t remove your hypotheses, so that didn’t need to be said. – Ted Shifrin Aug 16 '22 at 23:36
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    @TedShifrin, If you meant what I wrote above, then it is trivial since they are on the line then $|xA|+|AB|+|By|=|xy|$ and since $|xA|>0, |By|>0$ and $|AB|=r$, then $r<|xy|$. – RFZ Aug 16 '22 at 23:36
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    OK, now deduce the general case. – Ted Shifrin Aug 16 '22 at 23:37
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    @TedShifrin, I knew this hint but can you give more hint how to deduce the general case? – RFZ Aug 16 '22 at 23:40
  • To generalize, try proving that the point in both circles that minimizes $|xy|$ must lay on the line between the centers of both circles. You could try analytically by setting up an inequality for when the coordinates of a point are in one of the circles. – ArthD21 Aug 16 '22 at 23:41
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    @MathAndPhysics, I think there should be a more easier way to solve this. – RFZ Aug 16 '22 at 23:42
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    Obviously, you should introduce points $x’$ and $y’$ on the line in a way that $|xA|\ge |x’A|$, etc. Now finish and post the answer with no more questions! – Ted Shifrin Aug 16 '22 at 23:43
  • @TedShifrin, now it makes sense! I think in order to obtain such point $x'$ we just need to consider the right angle triangle with hypotenuse $xA$ and one side lying on a diameter of a circle right? – RFZ Aug 16 '22 at 23:45
  • Yes. Write up the complete proof :) – Ted Shifrin Aug 16 '22 at 23:48

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