# Talk:Descartes' theorem

I can't help wondering of Descartes' theorem should be a disambiguation page?  Thoughts? Michael Hardy 22:35, 23 May 2005 (UTC)

Is "curvature" the right term for ±1/r? I've seen explanations with curvature as always positive, 1/r, and "bend" used for the positive or negative curvature. Daggerbox (talk) 14:29, 13 February 2008 (UTC)

This article takes entirely too long before it states even the gist of Descartes' Theorem.  Ideally, at least the gist should appear in the lede, and the full statement should have been given by the next section (if not within the lede). A reader should not, for example, have first to read of the history of the theorem. —SlamDiego←T 03:26, 2 November 2008 (UTC)

The article mentions how the theorem has been extended to spheres and even arbitrary dimension hyperspheres. Is it possible anyone can add the equations for either of those? Even just 3D spheres for now would be nice.  —Preceding unsigned comment added by Skytopia (talk • contribs) 06:03, 26 February 2009 (UTC) 
It says there is no 3-dimensional analogue of the complex numbers that the higher-dimensional cases can use.  What about quaternions in a 3-dimensional case?

Testing the combination k_i = 1 1 1 gives  3 - 2*Sqrt(3) for the inner circle, which is negative. Is there supposed to be an initial +-? 121.45.179.128 (talk) 08:23, 16 December 2009 (UTC)

It is outer circe. Inner will be 3+2*Sqrt(3). You can also look at wikibooks article. --Adam majewski (talk) 18:04, 15 August 2010 (UTC)

In par 3 under Special Cases:
"...as it is not possible for three lines and one circle to be mutually tangent." - my immediate reaction was 'incircle of a triangle?' Probably should be clarified that mutual tangency of three lines makes no sense. But then the previous case of two parallel lines being considered mutually tangent to a pair of circles also becomes questionable, as tangency implies having common points. Thoughts?  —Preceding unsigned comment added by SwiftSurge (talk • contribs) 16:02, 13 August 2010 (UTC) 

It's not really true that "mutual tangency of three lines makes no sense". In the geometry of circles and lines it makes sense to think of lines as a limiting case of circles with infinite radius, and to think of parallel lines as being a limiting case of tangent circles. For instance, a Möbius transformation can transform tangent circles into parallel lines or vice versa. So with these conventions it is indeed possible for three lines to be mutually tangent: just make them all parallel. However in this case it would only be possible to make the fourth thing that is tangent to all of them be another line, not a circle. —David Eppstein (talk) 00:58, 14 August 2010 (UTC)

There seems to be an implicit assertion that given three mutually tangent circles, there always exists a fourth—but this isn't clearly spelled out.  Given a formula such as 
  
    
      
        
          k
          
            4
          
        
        =
        
          k
          
            1
          
        
        +
        
          k
          
            2
          
        
        +
        
          k
          
            3
          
        
        ±
        2
        
          
            
              k
              
                1
              
            
            
              k
              
                2
              
            
            +
            
              k
              
                2
              
            
            
              k
              
                3
              
            
            +
            
              k
              
                3
              
            
            
              k
              
                1
              
            
          
        
        .
        
      
    
    {\displaystyle k_{4}=k_{1}+k_{2}+k_{3}\pm 2{\sqrt {k_{1}k_{2}+k_{2}k_{3}+k_{3}k_{1}}}.\,}
  
, it's reasonable to ask if the expression under the square root can be negative.  Certainly if 
  
    
      
        
          k
          
            3
          
        
      
    
    {\displaystyle k_{3}}
  
 is zero, then the other circles must be external (you can't have a straight line fitting inside a circle), so 
  
    
      
        
          k
          
            1
          
        
      
    
    {\displaystyle k_{1}}
  
 and 
  
    
      
        
          k
          
            2
          
        
      
    
    {\displaystyle k_{2}}
  
 will be positive.  And if 
  
    
      
        
          k
          
            3
          
        
      
    
    {\displaystyle k_{3}}
  
 is strictly negative, the first two circles must be small enough to fit inside the third, so 
  
    
      
        
          k
          
            1
          
        
      
    
    {\displaystyle k_{1}}
  
 and 
  
    
      
        
          k
          
            2
          
        
      
    
    {\displaystyle k_{2}}
  
 are both positive and bigger in absolute value than 
  
    
      
        
          k
          
            3
          
        
      
    
    {\displaystyle k_{3}}
  
.  So I think Descartes' theorem does imply existence of a fourth circle.  But what I'm doing here is original research; is there a reference that clearly deals with the existence question? Jowa fan (talk) 04:19, 24 June 2011 (UTC)

It's a lot easier than that if you ignore the formula and think about it more geometrically. Perform an inversion centered on one of the points of tangency; the resulting configuration will consist of two parallel lines with a circle sandwiched between them, at which point it should be obvious that there are two choices for the fourth circle. Actually, according to Problem of Apollonius#Mutually tangent given circles: Soddy's circles and Descartes' theorem there are three more solutions: each of the given three circles is itself mutually tangent to all three circles as well. So anyway, yes, existence is known. —David Eppstein (talk) 06:48, 24 June 2011 (UTC)
That's all good info; this article should include enough of it to say that "such a circle must exist" and link where User:David Eppstein mentions.    yoyo (talk) 02:54, 25 January 2021 (UTC)

In its introduction the article refers to kissing circles.  Yet the standard definition of kissing, i.e. Osculating circle, implies 2nd order contact.  This is one order beyond tangency.  Two osculating circles would be the same circle, since their curvature would be the same. It seems kissing should be removed and only mutual tangency should be given as a condition.  — Preceding unsigned comment added by 66.188.89.180 (talk) 19:50, 20 March 2014 (UTC) 

They are certainly not osculating circles. But Soddy used the word "kissing" and for that reason I believe it should remain in the article. —David Eppstein (talk) 16:07, 21 March 2014 (UTC)

I've been writing a program to generate Apollonian Gaskets, and I've come to a minor hiccup.  It seems the ± sign in Equation 2 does not correspond to the ± sign in the unnumbered equation immediately after Equation 4 (Complex Descartes' Theorem).  In some cases, using a + sign in both equations yields one solution while using a − sign in both equations yields the other.  In other cases, both of those are wrong; one solution is found using a + sign in one equation and a − sign in the other, while the other solution is found by swapping those signs.  I haven't yet determined what makes the difference.  Vid the Kid (t/c) Yeah, that guy. 21:03, 18 March 2015 (UTC)
Thinking more about it, I suppose the ± in the expression for z4 is really just an extension of the problem of choosing a square root of a complex number.  To rephrase my original problem more aptly, I haven't found a way to match a solution for k4 with the corresponding solution for z4; the convention of the "principal root" being the one with a positive real part doesn't seem to help.  Although checking for geometric tangency seems the straightforward answer, I still hope to discover a more elementary test.  Vid the Kid (t/c) Yeah, that guy. 22:49, 18 March 2015 (UTC)

I came here having the same problem. After reading https://arxiv.org/abs/math/0101066, which was already cited in the sources for the complex theorem, I came up with a simple solution:
The paper features the equation 
  
    
      
        
          ∑
          
            j
            =
            1
          
          
            4
          
        
        
          b
          
            j
          
        
        (
        
          b
          
            j
          
        
        
          z
          
            j
          
        
        )
        =
        
          
            1
            2
          
        
        
          
            (
          
        
        
          ∑
          
            j
            =
            1
          
          
            4
          
        
        
          b
          
            j
          
        
        
          
            )
          
        
        
          
            (
          
        
        
          ∑
          
            j
            =
            1
          
          
            4
          
        
        
          b
          
            j
          
        
        
          z
          
            j
          
        
        
          
            )
          
        
        
        (
        2.3
        )
        ,
      
    
    {\displaystyle \sum _{j=1}^{4}b_{j}(b_{j}z_{j})={\frac {1}{2}}{\biggl (}\sum _{j=1}^{4}b_{j}{\biggr )}{\biggl (}\sum _{j=1}^{4}b_{j}z_{j}{\biggr )}\qquad (2.3),}
  
 which is given as a direct result of the complex descartes theorem. The exact way in which the paper derives this equation confuses me, but taking it for granted, it becomes clear how, using simple transformations, you can solve the equation for a unique value of 
  
    
      
        
          z
          
            4
          
        
      
    
    {\displaystyle z_{4}}
  
, namely 
  
    
      
        
          z
          
            4
          
        
        =
        
          
            
              −
              
                ∑
                
                  j
                  =
                  1
                
                
                  3
                
              
              
                b
                
                  j
                
              
              
                z
                
                  j
                
              
              
                
                  (
                
              
              
                
                  1
                  2
                
              
              
                
                  (
                
              
              
                ∑
                
                  k
                  =
                  1
                
                
                  4
                
              
              
                b
                
                  k
                
              
              
                
                  )
                
              
              −
              
                b
                
                  j
                
              
              
                
                  )
                
              
            
            
              
                b
                
                  4
                
              
              
                
                  (
                
              
              
                
                  1
                  2
                
              
              
                
                  (
                
              
              
                ∑
                
                  k
                  =
                  1
                
                
                  4
                
              
              
                b
                
                  k
                
              
              
                
                  )
                
              
              −
              
                b
                
                  4
                
              
              
                
                  )
                
              
            
          
        
        .
      
    
    {\displaystyle z_{4}={\frac {-\sum _{j=1}^{3}b_{j}z_{j}{\bigl (}{\frac {1}{2}}{\bigl (}\sum _{k=1}^{4}b_{k}{\bigr )}-b_{j}{\bigr )}}{b_{4}{\bigl (}{\frac {1}{2}}{\bigl (}\sum _{k=1}^{4}b_{k}{\bigr )}-b_{4}{\bigr )}}}.}
  

This formula works and gives one circle center for each of the two possible curvatures 
  
    
      
        
          b
          
            4
          
        
      
    
    {\displaystyle b_{4}}
  
 that were calculated using the regular Descartes theorem.
An exception is posed by the case of 
  
    
      
        
          
            
              1
              2
            
          
          
            
              (
            
          
          
            ∑
            
              k
              =
              1
            
            
              4
            
          
          
            b
            
              k
            
          
          
            
              )
            
          
          −
          
            b
            
              4
            
          
          =
          0
          
          ⟺
          
          

          
        
      
    
    {\displaystyle \textstyle {\frac {1}{2}}{\bigl (}\sum _{k=1}^{4}b_{k}{\bigr )}-b_{4}=0\iff {}}
  

  
    
      
        
          b
          
            4
          
        
        =
        
          b
          
            1
          
        
        +
        
          b
          
            2
          
        
        +
        
          b
          
            3
          
        
      
    
    {\displaystyle b_{4}=b_{1}+b_{2}+b_{3}}
  
, in which case the denominator becomes 0 and my formula no longer applies. For this case, one may recognize that this failure criterion is exactly the solution for 
  
    
      
        
          b
          
            4
          
        
      
    
    {\displaystyle b_{4}}
  
 from the regular Descartes theorem in the case where the discriminant is 0. In other words, the failure condition is exactly the case where you get a duplicate curvature for both possible tangent circles, and can't calculate a unique circle center from the curvature.
This also means that in the case where my formula doesn't apply, you can use the quadratic formula on the complex Descartes theorem to get two different circle centers from just one curvature, and this gives you both of the solutions.
So to summarize:
1. Use the quadratic formula on Descartes' theorem to find the two different values that the 4th circle's curvature 
  
    
      
        
          b
          
            4
          
        
      
    
    {\displaystyle b_{4}}
  
 can have.
2. If both values were the same because the discriminant is 0, use the quadratic formula on the complex Descartes' theorem to find the two different centers that the 4th circle can have.
3. If you instead found two different curvatures in step 1, use my formula to calculate one circle center corresponding to each curvature.
I also want to note that this method is only useful when you only have 3 circles at the start. At any moment where you have 4 circles, you can use simple addition and subtraction to reflect one of them over the other three to get a second circle that is tangent to those other three. The particular equations for this are given in the paper I linked at the start, but are again simple to derive from the (complex) Descartes theorem. This is simpler to implement not only because the math is easier, but also because it gives one unique next circle, as opposed to giving two different circles of which one might have been previously explored.
In case there is any confusion about the specific implementation, here is a demonstration that I wrote: https://github.com/HalbFettKaese/apollonian-circles
In its current state, the main page seems to suggest always using the quadratic formula to determine some combination of curvature and circle center selected by some unknown or inefficient manner. (for the latter, see Coding Train's implementation on the Apollonian Gasket, which seems to rely on the current main page's solution)
For this reason, I would also be interested in having some of this information added to the main page, but I don't have enough experience with editing pages to judge just how much of my explanation (or of a paraphrased version of it) still fits within the scope of the main page. HalbFettKaese (talk) 01:26, 23 September 2025 (UTC)

An external resource bears the title "Descartes' circle theorem".  
It's one thing to follow historical norms (however horrible), it's another thing to have page titles that are immediately self-evident.  
In this matter, for myself. I'd choose the later.  Good grief, Descartes' theorem has it's own disambiguation page.  How sucky is that?  If he'd been any more productive, perhaps the thing to do would be to assign his theorems Bach-Werke-Verzeichnis numbers.  At least then we could tell them apart on surface inspection.  — MaxEnt 18:45, 15 January 2017 (UTC)

Is it error? 
May be that for every three kissing...? Jumpow (talk) 07:33, 19 February 2018 (UTC)

No. The quadratic equation is the one labeled (1). It involves the radii of four circles, not three. —David Eppstein (talk) 07:51, 19 February 2018 (UTC)
Yes, of course. It was request from one of russian users and I did not recognize answer immeiately... Jumpow (talk) 13:31, 5 March 2018 (UTC)
It looks like an error: you start with three circles, solve the equation and reach a fourth. The way it's stated now reads like "you start with four circles, solve and equation, and reach a fourth circle", which doesn't make sense.Fernandohbc (talk) 02:41, 14 October 2018 (UTC)
Answering myself: After re-reading, I realize that my paraphrasing is wrong too. My apologies. I think that the clarification needs to be made in the following sentence, though. Something along the lines of "By solving this equation, one can start with three given mutually tangent circles and construct a fourth circle tangent to all of them.", maybe?Fernandohbc (talk) 02:46, 14 October 2018 (UTC)
If readers don't understand that an equation involving four things can be solved to find one of the things from the other three, then perhaps their level of mathematical sophistication does not reach that of the target audience for this article. —David Eppstein (talk) 03:26, 14 October 2018 (UTC)
This is not what I claimed readers would misunderstand. My comment was regarding the phrasing in plain English, which is a bit confusing, not the mathematical sophistication of the theorem itself. Fernandohbc (talk) 04:01, 14 October 2018 (UTC)

The article could usefully tell readers where they may find proofs of the given facts, solutions and equations.  Not implying that the article needs to contain proofs, since Wikipedia doesn't do that.  yoyo (talk) 02:58, 25 January 2021 (UTC)

I agree with this. An article about a theorem should do a bit better job showing or at least explicitly linking to proofs. The history section mentions that "Multiple proofs of the theorem have been published" but does not directly cite any of those inline, and the few-words summaries are unhelpfully vague. The "Proof from Heron's formula" is algebraic and not especially satisfying. @David Eppstein have you done any survey of the published proofs of Descartes' theorem? Can we make a section including a couple of them or at least providing a sketch? I feel like a section about proofs of this theorem (and maybe some more images for the history section) is the only thing really missing to make this "good article" quality. –jacobolus (t) 18:15, 8 July 2023 (UTC)
To be honest, the Heron's formula proof is the least algebraic and unsatisfying proof that I encountered. They all seem to have a similar flavor: take a relatively simple piece of geometry (Heron's formula, Pappus chains, etc) and then elaborate it into a page or two of algebra to get the formula. —David Eppstein (talk) 18:32, 8 July 2023 (UTC)
That's a bummer. We should perhaps at least track down some citations to the original sources then, so readers can save some effort on their path to disappointment. :-) –jacobolus (t) 18:59, 8 July 2023 (UTC)
any updates on this? WorldDiagram837 (talk) 01:00, 13 February 2025 (UTC)
Feel free to go look in old sources to see if you can find any nice proof(s). –jacobolus (t) 05:46, 13 February 2025 (UTC)
But did you notice the proof in § Soddy circles of a triangle? –jacobolus (t) 05:50, 13 February 2025 (UTC)
uh I did. Any problems in that or something or are you telling me the fact that a proof exists on the article page?(I think the proof was being discussed in the talk topics below this) WorldDiagram837 (talk) 13:07, 13 February 2025 (UTC)
oh right you were the one who proposed the Heron's formula proof. my bad. Yes, as a matter of fact I did notice that. Imma read that now WorldDiagram837 (talk) 13:13, 13 February 2025 (UTC)

It might be nice to add a section about more general relationships between circles, beyond Steiner chains. For example here's a paper where out of four circles two of them intersect, with the other two tangent to both of those and each-other:

Fox, M. D. (1980), "Formulae for the curvatures of circles in chains", The American Mathematical Monthly, 87 (9): 708–715, doi:10.1080/00029890.1980.11995129, JSTOR 2321856
And here is a paper with a generalization to arbitrary circles in the plane:

Kocik, Jerzy (2010), "Golden window" (PDF), Mathematics Magazine, 83 (5): 384–390, doi:10.4169/002557010X529815
–jacobolus (t) 07:02, 11 June 2023 (UTC)

Isn't that really more suitable for our article on the Problem of Apollonius, which is exactly about finding tangent circles to triples of circles that are not necessarily tangent? Or, for constructing patterns of tangent circles with arbitrary planar graphs of adjacencies (like the golden window example), see circle packing theorem. —David Eppstein (talk) 07:17, 11 June 2023 (UTC)
Let me be more precise. Lagarias/Mallows/Wilks found that Descartes' theorem could be expressed as a matrix equation 
  
    
      
        
          k
          
            T
          
        
        Q
        k
        =
        0
        ,
      
    
    {\displaystyle k^{T}Qk=0,}
  
 where 
  
    
      
        k
      
    
    {\displaystyle k}
  
 is the vector of curvatures of the circles and

  
    
      
        Q
        =
        
          
            [
            
              
                
                  
                    
                      
                        −
                      
                    
                  
                  1
                
                
                  −
                  1
                
                
                  −
                  1
                
                
                  −
                  1
                
              
              
                
                  −
                  1
                
                
                  
                    
                      
                        −
                      
                    
                  
                  1
                
                
                  −
                  1
                
                
                  −
                  1
                
              
              
                
                  −
                  1
                
                
                  −
                  1
                
                
                  
                    
                      
                        −
                      
                    
                  
                  1
                
                
                  −
                  1
                
              
              
                
                  −
                  1
                
                
                  −
                  1
                
                
                  −
                  1
                
                
                  
                    
                      
                        −
                      
                    
                  
                  1
                
              
            
            ]
          
        
        .
      
    
    {\displaystyle Q={\begin{bmatrix}{\phantom {-}}1&-1&-1&-1\\-1&{\phantom {-}}1&-1&-1\\-1&-1&{\phantom {-}}1&-1\\-1&-1&-1&{\phantom {-}}1\\\end{bmatrix}}.}
  

What Kocik's paper claims is that these coefficients come from inverting a matrix 
  
    
      
        
          Q
          
            −
            1
          
        
      
    
    {\displaystyle Q^{-1}}
  
 whose entries are the inversive distances of each pairs of circles in the resulting configuration (including of each circle with itself along the diagonal), and by changing the coefficients in the matrix we can enforce different conditions on the circles. (I got tripped up a bit because the particular "Descartes configuration" matrix above is a scalar multiple of its inverse.)
(I guess with the signs I wrote above, these coefficients are the negatives of inversive distance as defined on Wikipedia. The cosine of the angle between between two oriented lines is 1 for parallel lines, 
  
    
      
        −
        1
      
    
    {\displaystyle -1}
  
 for lines pointed in diametrically opposite directions, or 
  
    
      
        0
      
    
    {\displaystyle 0}
  
 for perpendicular lines. The negative inversive distance is a generalization which applies to oriented circles, and is the cosine of the angle between circles which intersect, or falls outside the interval 
  
    
      
        [
        −
        1
        ,
        1
        ]
      
    
    {\displaystyle [-1,1]}
  
 for circles which do not intersect.)
Kocik has some other papers with more detail, e.g.
Kocik, Jerzy (2007), A theorem on circle configurations, arXiv:0706.0372
Kocik, Jerzy (2019), Proof of Descartes circle formula and its generalization clarified, arXiv:1910.09174
–jacobolus (t) 00:33, 12 June 2023 (UTC)
FWIW the Steiner chain article has an incompatible definition of inversive distance using logs. The one I used here is the same as our inversive distance article (and the source I used), but that confused me for a while. —David Eppstein (talk) 01:59, 12 June 2023 (UTC)
The general concept surely dates from the 19th century. Who came up with the version in the Wikipedia article? I think they messed up with the name; it could be called the "inversive circle dot product" or "circle cosine" or something (Kocik uses the name "Pedoe product"), as even though distance is involved it's not properly a type of "distance" per se. I think they also messed up the sign; the product of a circle with itself should be 1 here.
Coxeter's version of "inversive distance" apparently only considers non-intersecting circles and takes the arccosh of the absolute value of the Wikipedia article's concept, which is more like a distance, but has the problem that it erases the orientation of the circle from consideration and also leaves out consideration of intersecting circles, both of which seem to me like mistakes.
A bit of a tangent here, while we're talking about oriented circles, I want to eventually write an article called something like Laguerre geometry or perhaps Line geometry that does a better job of describing Edmond Laguerre's dual approach to looking at Euclidean geometry by starting from oriented lines as the primitive object rather than points. I collected a long list of relevant sources at Talk:Laguerre transformations § List of references but figuring out what the scope and focus should be for an article and how to organize it is a bit daunting, since this is a topic which appears regularly here and there but is uncommon enough to have never quite crystallized into part of the canon or have particularly widely adopted or settled conventions. Any suggestions? –jacobolus (t) 02:38, 12 June 2023 (UTC)
Oriented projective geometry is symmetric with respect to point-line duality, so what difference does it make which ones you call the primitive objects? Or is this some kind of geometry that is not projective? —David Eppstein (talk) 03:45, 12 June 2023 (UTC)
We are talking about a kind of dual metrical geometry, not just projective relationships. If you take lines to be primitives there is a whole set of transformations dual to Möbius transformations, called Laguerre transformations, which preserve line–circle tangencies and preserve distances along lines between them but do not preserve points (for example two intersecting circles might map to two non-intersecting circles under a Laguerre transformation, but the pair of lines each tangent to both original circles will map to a pair of lines tangent to both resulting circles).
On the sphere, where both distances and angles are circular, there's a precise duality between the two (the dual of every point is an oriented great circle, and vice versa; every triangle has a dual "trilateral", every spherical conic as a locus of points has a dual spherical conic which is the envelope of lines, and so on), so spherical Laguerre transformations can be represented, just like spherical Möbius transformations, as fractional linear transformations of complex numbers; but the Euclidean plane has circular angles and flat distances, so the dual Laguerre geometry switches these, and Laguerre transformations are modeled by fractional linear transformations of dual numbers. In the hyperbolic plane, Laguerre transformations are represented by fractional linear transformations of hyperbolic numbers.
There are some nifty solutions to Euclidean problems using Laguerre geometry. For instance, given 3 oriented circles there are only 2 solutions to the problem of Apollonius, and you can find them by applying a transformation that turns all three of your original circles into points, taking the two circles passing through those points with opposite directions, and invert the transformation to recover the two circles you are looking for. (If you want to find all 8 solutions to the un-oriented circle problem, you have 4 choices for the relative orientations of the 3 source circles.)
Pretty well every feature of point-based metrical geometry has an oriented-line-based dual.
As an example of a practical application, a "circular mesh" is a quadrilateral mesh where every face is a cyclic quadrilateral, a property preserved under Möbius transformations. A dual idea, the "edge offset mesh" has every edge emanating from a vertex lying on a common right circular cone, a property preserved under Laguerre transformations. http://fractal.dam.fmph.uniba.sk/~chalmo/GM/edge.pdf
–jacobolus (t) 05:29, 12 June 2023 (UTC)
Oh, from looking at that list again, Coolidge has the sign right (p. 109, p. 351, p. 408) for the cosine of the angle between circles. –jacobolus (t) 02:55, 12 June 2023 (UTC)
I'm thinking of adding something after the Steiner chain section along the lines of the following. Thoughts? –jacobolus (t) 20:51, 12 June 2023 (UTC)

Removing this draft subsection and moving it to the article. –jacobolus (t) 22:05, 15 June 2023 (UTC)
I think it's helpful to directly show the matrix 
  
    
      
        
          
            Q
          
          
            −
            1
          
        
      
    
    {\displaystyle \mathbf {Q} ^{-1}}
  
 and vector 
  
    
      
        
          k
        
        ,
      
    
    {\displaystyle \mathbf {k} ,}
  
 because then readers can very easily see how this matrix equation is equivalent to "equation 1" from earlier in the article. It seems to me that taking that part out makes that section of the article slightly more compact but substantially harder for semi-technical readers to follow (say someone a few years out of college who only ever took one introductory linear algebra course), because the section then becomes much less concrete. –jacobolus (t) 03:19, 19 June 2023 (UTC)


### special case of "steiner chains"
@David Eppstein – the whole section you just removed is a special case of the generalization of Descartes theorem. The discussion was focused on the algebraic identity which is directly analogous to the "Descartes theorem" identity for the mutually-tangent-circle configuration. The purpose of including it was to (a) demonstrate how the generalization works in a specific case, and (b) show how the two "soddy circles" (related by the ± cases of the find-the-fourth-curvature version of Descartes' theorem) are strictly related to each-other as well as related to the other three circles, while (c) also showing how Descartes' theorem is related to nearby topics in inversive geometry. All of that is directly relevant to this article. I don't see how removing it benefits readers. We don't have a page limit here, but even if we did this article is not excessively long or sprawling. Many (most?) readers probably won't read all the way from beginning to end, but it's not like they're being subjected to an arbitrary snorefest mishmash here, and some readers will likely find the section relevant and interesting. –jacobolus (t) 01:58, 20 July 2023 (UTC)

To me the whole article felt unfocused; not something I would feel comfortable nominating for GA status, particularly because of WP:GACR #3b, requiring good articles to stay focused on their main topic and not go into unnecessary detail. This article had started to spread out into lots of topics that are not actually the formula for curvature for four mutually tangent circles. Steiner's porism is one of them. We have an article about it; we don't need to replicate it here. Inversive distance is another. Again, we have an article about it, where more detailed material can go. The fact that you can measure the inversive distances of arbitrary circles and put them into a matrix is not particularly informative. And again, it's not particularly on-topic here. In that sense, the sub-sub-section I removed is a tangent to a tangent.
More generally, there is certainly a benefit to the reader in not distracting them with irrelevancies. A particularly bad example in that respect is Chernoff bound. It's a topic that is often useful in algorithms research, and so I would like to refer to our article when I want to look up the precise form of the bound. The form I want is always one of the forms in sub-subsection Chernoff bound § Multiplicative form (relative error). All the rest of the article is filler. Because the part I want is buried two levels down somewhere in the middle of the article, it always takes me much longer to find it than it would for a better-focused article. All the rest is literally wasting my time, repeatedly, every time I want to refer to the article. It's annoying. I don't want this article to be equally annoying by becoming buried in stuff you don't want to look up to the point where finding the parts you do want to look up becomes difficult. —David Eppstein (talk) 05:00, 20 July 2023 (UTC)
The problem is that not every reader has quite as much experience or context as you have. So parts that seem to you to be unnecessary filler may be essential context for someone else. (This is of course a universal challenge in technical writing.)
Thanks for at least creating Soddy circles of a triangle instead of just deleting that part; we should probably merge Soddy line into there, and that does give the subject a bit more room to expand without getting out of scope here.
We should add an image or two back to the relevant section of this article. Any advice on what it should include?
I wonder if we can rewrite the "proof" part to be a bit clearer (whether by reorganizing, adding a picture, chopping some parts, ...?) I find my eyes glaze over unless I really try to pay attention to it. –jacobolus (t) 07:22, 20 July 2023 (UTC)
Re image: Something less busy than the ones now in Soddy circles of a triangle, I think, but that still identifies the variables used in that section.
Re merge between Soddy circles of a triangle and Soddy line: yes, this looks reasonable to me.
I would have no objection to similarly repurposing the removed subsection on Steiner chains as part of another article. I don't think it's bad content, in general. It just seemed overdetailed regarding stuff that is not the relation between curvatures of four tangent circles for inclusion in this specific article. —David Eppstein (talk) 18:02, 20 July 2023 (UTC)
Also are there any images we can find for the history section? –jacobolus (t) 07:25, 20 July 2023 (UTC)
If we can find facsimiles of Descartes' original letter, Yamaji's work, or the 1824 Sangaku, those would fit. —David Eppstein (talk) 18:03, 20 July 2023 (UTC)

I'm not able to edit the article right now due to Wikipedia technical problems (keeps timing out), but I made an image showing the 4 radii in the Heron's formula proof. If I can't successfully add it to the article, feel free to. –jacobolus (t) 00:20, 21 July 2023 (UTC)

I wonder if it would be better to show the whole sides of the triangles labeled 
  
    
      
        
          r
          
            1
          
        
        +
        
          r
          
            2
          
        
      
    
    {\displaystyle r_{1}+r_{2}}
  
 etc. instead of splitting each side up into two parts and labeling them separately. –jacobolus (t) 00:29, 21 July 2023 (UTC)
Okay I made that change, which I think makes the picture a bit more legible. –jacobolus (t) 00:46, 21 July 2023 (UTC)
New picture looks helpful to me. —David Eppstein (talk) 07:28, 21 July 2023 (UTC)
@David Eppstein: maybe the negative radii are too tricksy, but I think signed areas are an improvement, because there are actually several possibilities here: (1) the 4th curvature is negative and 
  
    
      
        
          P
          
            4
          
        
      
    
    {\displaystyle P_{4}}
  
 is inside 
  
    
      
        △
        
          P
          
            1
          
        
        
          P
          
            2
          
        
        
          P
          
            3
          
        
        ,
      
    
    {\displaystyle \triangle P_{1}P_{2}P_{3},}
  
 (2) the curvature is negative and 
  
    
      
        
          P
          
            4
          
        
      
    
    {\displaystyle P_{4}}
  
 is in the inside of one angle of 
  
    
      
        △
        
          P
          
            1
          
        
        
          P
          
            2
          
        
        
          P
          
            3
          
        
      
    
    {\displaystyle \triangle P_{1}P_{2}P_{3}}
  
 but outside the triangle across the opposite side, (3) 
  
    
      
        
          P
          
            4
          
        
      
    
    {\displaystyle P_{4}}
  
 is at infinity, (4) the curvature is positive and 
  
    
      
        
          P
          
            4
          
        
      
    
    {\displaystyle P_{4}}
  
 is outside all three angles of the triangle (in the interior of one of the vertical angles. The easiest way to deal with all of these is to allow a negative radius, but it might be unnecessarily confusing. The next easiest way is to note that the sides adjacent to 
  
    
      
        
          P
          
            4
          
        
      
    
    {\displaystyle P_{4}}
  
 are differences of radii in cases (1–2), but to allow signed areas so that the sum 
  
    
      
        
          Δ
          
            123
          
        
        =
        
          Δ
          
            423
          
        
        +
        
          Δ
          
            143
          
        
        +
        
          Δ
          
            124
          
        
      
    
    {\displaystyle \Delta _{123}=\Delta _{423}+\Delta _{143}+\Delta _{124}}
  
 still holds. The most complicated is to treat every case separately. In case (1) the formula is 
  
    
      
        
          Δ
          
            123
          
        
        =
        
          Δ
          
            423
          
        
        +
        
          Δ
          
            143
          
        
        +
        
          Δ
          
            124
          
        
      
    
    {\displaystyle \Delta _{123}=\Delta _{423}+\Delta _{143}+\Delta _{124}}
  
 in case (2) e.g. 
  
    
      
        
          Δ
          
            123
          
        
        +
        
          Δ
          
            423
          
        
        =
        
          Δ
          
            143
          
        
        +
        
          Δ
          
            124
          
        
      
    
    {\displaystyle \Delta _{123}+\Delta _{423}=\Delta _{143}+\Delta _{124}}
  
 and in case (4) e.g. 
  
    
      
        
          Δ
          
            123
          
        
        +
        
          Δ
          
            423
          
        
        +
        
          Δ
          
            143
          
        
        =
        
          Δ
          
            124
          
        
      
    
    {\displaystyle \Delta _{123}+\Delta _{423}+\Delta _{143}=\Delta _{124}}
  
 –jacobolus (t) 20:10, 21 July 2023 (UTC)

I think this article should be readable by people who don't care to understand signed areas. And they are completely unnecessary here. See WP:TECHNICAL (which by the way is one of the GA requirements): We should not make the article more technical than it needs to be. Signed areas are an unnecessary technicality. As for P_4 at infinity: let's not get into trying to set up equations of equal area of triangles that have a vertex at infinity. That way lies original research and dubious definitions and confusion. —David Eppstein (talk) 20:23, 21 July 2023 (UTC)
To be clear: I am not suggesting we should try to get fancy about areas in the straight line case. It just falls between the other two (I guess I might also mention one additional case when the center is on the triangle and one of the triangles degenerates, with area 0). Maybe the clearest for this proof is to just enumerate the possibilities then. –jacobolus (t) 20:48, 21 July 2023 (UTC)
I guess your alternative is okay. Let me read more carefully. –jacobolus (t) 20:49, 21 July 2023 (UTC)

Is there a solution (straightedge and compass construction) using any right-angled triangle without first having to calculate the position of the Descartes circles? Petrus3743 (talk) 21:49, 13 December 2023 (UTC)

Meaning, is there a straightedge and compass construction of the fourth circle tangent to three given circles? Yes, but this is a different topic (and a much older one) than Descartes' theorem. It is the Problem of Apollonius, which we describe early in the history of this section. It is the attempted reformulation of this problem using algebra rather than compass and straightedge that led Descartes to Descartes' theorem, as we also describe in the history section. —David Eppstein (talk) 21:59, 13 December 2023 (UTC)
Thanks for your quick answer. The question was, is there a construction with any right-angled triangle without first calculating the position of the 4th circle?--Petrus3743 (talk) 22:21, 13 December 2023 (UTC)
A construction of what, from what givens? If you are just going to make me guess what you mean by repeating the same question in the same words that you already asked, I am not going to be able to give you a better answer than the one I already gave. —David Eppstein (talk) 02:08, 14 December 2023 (UTC)
Maybe the following picture is better than my bad English. Is there a reference for this: GeoGebra? — Preceding unsigned comment added by Petrus3743 (talk • contribs) 06:04, 14 December 2023 (UTC)
Yes, see Problem of Apollonius. This is a particular special case. The three circles at the vertices have diameters 
  
    
      
        −
        a
        +
        b
        +
        c
        ,
      
    
    {\displaystyle -a+b+c,}
  
 
  
    
      
        a
        −
        b
        +
        c
        ,
      
    
    {\displaystyle a-b+c,}
  
 and 
  
    
      
        a
        +
        b
        −
        c
        ,
      
    
    {\displaystyle a+b-c,}
  
 (for a triangle of sides 
  
    
      
        a
        ,
        b
        ,
        c
      
    
    {\displaystyle a,b,c}
  
), so those are not hard to construct. –jacobolus (t) 06:36, 14 December 2023 (UTC)
The special case in which the three initial circles are tangent, as in the GeoGebra link, is easier than the general problem. See my 1999 web page https://ics.uci.edu/~eppstein/junkyard/tangencies/apollonian.html and related paper "Tangent Spheres and Triangle Centers" (Amer. Math. Monthly 2001). The right angle helps a tiny bit more, as two of the first three lines of the construction are already sides of the right triangle. —David Eppstein (talk) 06:42, 14 December 2023 (UTC)
Thanks for the good solution! Maybe I will also draw your construction and enter it into the article Apollonisches Problem with the two links as references.--Petrus3743 (talk) 15:19, 14 December 2023 (UTC)
Thank you again for your references! I was able to use it well in the article Satz von Descartes. With regards--Petrus3743 (talk) 11:46, 18 December 2023 (UTC)

...which has been described as "a sort of demented version of the Pythagorean theorem".
and what will the reader be able to get from this? (I dont understand it myself, hence I am asking)
(located in the Special Cases section) WorldDiagram837 (talk) 01:03, 13 February 2025 (UTC)

Instead of 
  
    
      
        
          a
          
            2
          
        
        +
        
          b
          
            2
          
        
        =
        
          c
          
            2
          
        
      
    
    {\displaystyle a^{2}+b^{2}=c^{2}}
  
 we have 
  
    
      
        
          a
          
            1
            
              /
            
            2
          
        
        +
        
          b
          
            1
            
              /
            
            2
          
        
        =
        
          c
          
            1
            
              /
            
            2
          
        
      
    
    {\displaystyle a^{1/2}+b^{1/2}=c^{1/2}}
  
. The comment is pointing out the syntactic similarity to the Pythagorean theorem while also noting that it doesn't actually mean the same thing. —David Eppstein (talk) 03:02, 13 February 2025 (UTC)
gotcha thanks for explaining  WorldDiagram837 (talk) 13:08, 13 February 2025 (UTC)
but isnt it root k4 = root k1 +/- root k2 (apologies I dont how to use the math language here)
what about the time when I use the - sign? It won't be the Pythagorean form then, will it? WorldDiagram837 (talk) 13:29, 13 February 2025 (UTC)
You could express the Pythagorean theorem with the ± sign too, or change which variables go on which side. It would still describe right triangles: a^2 - b^2 = c^2 with a minus is the same as a^2 = b^2 + c^2 with a as the hypotenuse of the triangle. So that's not really a way that these equations differ. —David Eppstein (talk) 18:04, 13 February 2025 (UTC)
alright thanks WorldDiagram837 (talk) 01:32, 14 February 2025 (UTC)