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This is hard to search and probably easy to solve, but I keep finding articles about intersecting circles, and that is not what I'm after. I don't know what to tag this under, so if you know how to classify this better, please do.

circles.

I'd like to know how the red circle's radius is related to the radii of the larger circles, and why.

AakashM
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user
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2 Answers2

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Your setup is a special case of Descartes' theorem. In general that theorem describes the radii of four circles which all touch one another, but in your special case one of these circles is a line. Since the key ingredient to the relation are inverse radii, this means that the corresponding term will simply be zero.

Long story short: if $r_1$ and $r_2$ are the radii of the given circles, then the small circle has radius

$$r_3 = \frac{1}{\dfrac1{r_1}+\dfrac1{r_2}+\dfrac2{\sqrt{r_1r_2}}}$$

The only idea for a proof which comes to my mind just now would involve Lie geometry, which might be a bit beyond the scope of this question. But perhaps referring to Descartes' theorem is proof enough here?

MvG
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  • There is an exterior circle tangent to the circles r1 ,r2 which lies on the horizontal line and which comes from taking the negative root of the expression rather than the positive root, this is the fourth circle in Descartes Theorem. – Alan Oct 18 '13 at 10:43
  • @Alan: I agree that there is an alternate solution which touches the circles from the outside, but that is not the fourth circle of the theorem, since the inner and the outer circle don't touch while all three touch the line. – MvG Oct 18 '13 at 12:48
  • Yes ,there is both an inner and an outer circle, thank you. – Alan Oct 18 '13 at 16:55
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Let $a$ & $b$ be the radii of the larger circles then the radius $r$ of the smaller circle (derived here) is given by the general expression $$\bbox [4pt, border: 1px solid blue;]{ \color{red}{r=\frac{ab}{a+b+2\sqrt{ab}}} }$$

Harish Chandra Rajpoot
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