D'Alembert theorem

If three circles $A$, $B$, and $C$ are taken in pairs, the external similarity points of the three pairs lie on a straight line ($I,J,K$ are aligned). Similarly, the external similarity point of one pair and the two internal similarity points of the other two pairs lie upon a straight line, forming a similarity axis of the three circles (Two inner centers are aligned with an outer center: $I',K',J$ or $I',J',K$ or $I,J',K'$).

[PDF] [TEX]

\documentclass{standalone} 
\usepackage{tkz-euclide}
\tkzSetUpPoint[size=1,color=teal]
\tkzSetUpLine[thin,color=teal]
\tkzSetUpCompass[color=orange,ultra thin,/tkzcompass/delta=10]
\tikzset{label style/.append style={color=teal}}
\tikzset{new/.style={color=orange,ultra thin}} 
\tikzset{step 1/.style={color=cyan,ultra thin}} 
\tikzset{step 2/.style={color=purple,ultra thin}}

\begin{document} 
  \begin{tikzpicture}
  \tkzDefPoints{0/0/A,3/0/a,7/-1/B,5.5/-1/b,5/-4/C,4.25/-4/c}
  \tkzDrawCircles[fill=teal!20,opacity=.3](A,a B,b C,c)
  \tkzDefExtSimilitudeCenter(A,a)(B,b)             \tkzGetPoint{I}
  \tkzDefExtSimilitudeCenter(A,a)(C,c)             \tkzGetPoint{J}
  \tkzDefExtSimilitudeCenter(C,c)(B,b)             \tkzGetPoint{K}
  \tkzDefIntSimilitudeCenter(A,a)(B,b)             \tkzGetPoint{I'}
  \tkzDefIntSimilitudeCenter(A,a)(C,c)             \tkzGetPoint{J'}
  \tkzDefIntSimilitudeCenter(C,c)(B,b)             \tkzGetPoint{K'}
  \tkzDrawPoints(A,B,C,I,J,K,I',J',K')
  \tkzDrawSegments[step 1](I,K A,I A,J B,I B,K C,J C,K)
  \tkzDrawSegments[step 2](I,J' I',J I',K)
  \tkzLabelPoints(I,J,K,I',J',K')
  \end{tikzpicture}
\end{document}