I drew a circle for Pappus chain on an arbelos. How to draw other circles?

enter image description here




\begin{tikzpicture}[scale=1.5,line join=round,font=\small]
\coordinate[label=below:$A$] (A) at (0,0);
\coordinate[label=below:$B$] (B) at (4,0);
\coordinate[label=below:$C$] (C) at (6,0);
(6,0) arc [radius=3, start angle=0, delta angle=180] -- 
(0,0) arc [radius=2, start angle=180, delta angle=-180]--(4,0) arc [radius=1, start angle=-180, delta angle=-180]--cycle;

\path[name path=b1] 
(6,0) arc [radius=3, start angle=0, delta angle=180];
\path[name path=b2] (0,0) arc [radius=2, start angle=180, delta angle=-180];
\path[name path=b3] (4,0) arc [radius=1, start angle=-180, delta angle=-180];

\draw[olive!50,name path=b4] (4,0) arc (225:60:{sqrt(2)});
\draw[violet!50,name path=b5] (4,4) arc (45:-40:{2*sqrt(2)});
\path [name intersections={of = b4 and b5, by={D}}];
\path [name intersections={of = b2 and b4, by={T,E}}];
\path [name intersections={of = b3 and b5, by={F}}];
\coordinate (K) at ($(E)!.5!(F)$);
\coordinate (L) at ($(D)!.5!(F)$);
\coordinate (M) at ($(K)!1cm!90:(F)$);
\coordinate (N) at ($(L)!1cm!90:(D)$);
\path[name path=g1] (K)--(M);
\path[name path=g2] (L)--(N);
\path [name intersections={of = g1 and g2, by={O}}];
  \p1=( $(E)-(O) $ )
  (O) circle ({veclen(\x1,\y1)});
\foreach \p in {D,E,F,O}
\draw[fill,DodgerBlue4] (\p) circle(1pt);
%\foreach \p in {D,E,F}
%\node[above] at (\p) {$\p$};
  • 2
    Adding a link to for example en.wikipedia.org/wiki/Pappus_chain might be useful, as a lot of (most?) people won't know what a Pappus chain is, and thus wouldn't understand what you want to do. Apr 17, 2017 at 6:05

3 Answers 3


With the coordinates provided in the links:


  \coordinate[label=below:A](A) at (0,0);
  \coordinate[label=below:B](B) at (\RadiiBig*2,0);
  \coordinate[label=below:C](C) at (\RadiiSmall*2,0);
  \draw(\RadiiBig,0) circle (\RadiiBig);
  \draw(\RadiiSmall,0) circle (\RadiiSmall);
  \draw ($(C)+(\RadiiBig-\RadiiSmall,0)$) circle (\RadiiBig-\RadiiSmall);
  \foreach \n in {1,...,15}{%
    \draw(\x,\y) circle (\Radn);
    \draw(\x,-\y) circle (\Radn);

enter image description here


Following Thruston's idea with circle inversion applied to Tikz. With use of the definition that a a point P is inverted to P' by AP*AP'=r^2 where r is the mirroring circle, I get the following. I did only a function for the inversion of a point, so the center of the mirror image of a circle has to be derived from that.

  declare function={CircInv(\APp,\r)=\r*\r/\APp;}
  \coordinate[label=below:A](A) at (0,0);
  \coordinate[label=below:B](B) at (\RadiiBig*2,0);
  \coordinate[label=below:C](C) at (\RadiiSmall*2,0);
  \pgfmathsetmacro{\Rp}{\RadiiBig-\RadiiSmall}%% R'
  \pgfmathsetmacro{\r}{2*\RadiiBig}%% r: mirror circle
  \pgfmathsetmacro{\R}{(CircInv(2*\RadiiSmall,\r)-\r)/2}%% R: Mirror image of first circle
  \coordinate[label=below:$P'$](Pp) at ($(C)+(\Rp,0)$);%% Center first circle
  \coordinate[label=below:$P$](P) at ($(B)+(\R,0)$);%% Center of mirror circle of first circle
  \draw(\RadiiBig,0) circle (\RadiiBig);
  \draw(\RadiiSmall,0) circle (\RadiiSmall);
  \draw[red](A) (-20:\r) arc (-20:50:\r);\draw[red](A) -- +(-15:\r) node[pos=0.5,anchor=south]{$r$};
  \draw[blue](Pp) circle (\Rp);\draw[blue] (Pp) -- +(45:\Rp) node[pos=0.5,anchor=south east]{$R'$};
  \draw[gray](P) circle (\R);\draw[gray] (P) -- +(45:\R) node[pos=0.5,anchor=south east]{$R$};
  \foreach \n in {1,2}{%% First loop showing the column circles
    \pgfmathsetmacro{\X}{2*\n*\R/sin(\alpha)}%% Length from (A) to center of circle
    \pgfmathsetmacro{\Rpn}{(CircInv(\X-\R,\r)-CircInv(\X+\R,\r))/2}%% Size of mirror circle
    \pgfmathsetmacro{\Y}{CircInv(\X+\R,\r)+\Rpn}%% Length (A to center mirror circle
    \draw[gray](A) -- (\alpha: \X);
    \draw[gray](P |- {(0,2*\n*\R)}) circle (\R);
    \draw[blue](A) -- (\alpha: \Y) circle (\Rpn);
  \foreach \n in {3,4,...,30}{%% Same as previous loop but without gray stuff
    \draw[blue](A) (\alpha: \Y) circle (\Rpn);    

enter image description here

  • Added a Tikz version of the solution by @Thruston
    – StefanH
    Apr 19, 2017 at 8:11

You can draw this with circle inversion as well.

enter image description here

I'm not sure how to do inversion with TikZ but here is a Metapost version, wrapped up in luamplib. Compile with lualatex.

% invert path or pair P in circle C
vardef invert(expr P, C) = 
  save I, r; pair I; numeric r;
  I = center C;
  r = abs(point 0 of C shifted -I); 
  if pair P: if abs(P-I) > 0: unitvector(P-I) scaled (r/abs(P-I)*r) shifted fi I
  elseif path P:
     save T; numeric T;
     T = length P;
     for t=0 upto T-1: invert(point t of P, C) .. endfor if cycle P: cycle else: invert(point T of P, C) fi

    pair A,B,C;
    numeric r; 

    A = origin;
    C = (10cm,0);

    r = 3/4;
    B = r[A,C];

    path c[];
    c1 = fullcircle scaled 2 abs(A-C); % large circle for the inversions
    c2 = fullcircle scaled abs(A-C) shifted 1/2[A,C];
    c3 = fullcircle scaled abs(A-B) shifted 1/2[A,B];
    c4 = fullcircle scaled abs(B-C) shifted 1/2[B,C];
    c5 = invert(c4,c1);

    numeric d; d = abs(point 0 of c5-point 4 of c5);

    for i=1 upto 42:
        draw invert(c5 shifted (0,i*d), c1);

    draw subpath(0,4) of c2 withcolor 2/3 blue;
    draw subpath(0,4) of c3 withcolor 2/3 blue;
    draw subpath(0,4) of c4 withcolor 2/3 blue;

    draw A--C;
    dotlabel.bot("$A$", A);
    dotlabel.bot("$B$", B);
    dotlabel.bot("$C$", C);


The method is a bit easier to understand if I draw in a few more parts of the construction: the grey circles are c5 in my code; the pink arc is part of c1, which is the circle in which the c5s in the column are inverted.

enter image description here

  • Nice solution. I have added a version based on Tikz in my answer.
    – StefanH
    Apr 19, 2017 at 8:10

Here a simple solution with tkz-euclide v5 the latest version of tkz-euclide: see section 47.25 of the manual.

\usepackage{tkz-euclide}% v5


  \foreach \i in {-\nc,...,0,...,\nc}
   \tkzDefCircleBy[inversion = center A through C](J,H)


enter image description here

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