<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\subsection{Definition of circle by transformation; \tkzcname{tkzDefCircleBy} }
These transformations are:

\begin{itemize}
   \item translation;
   \item homothety;
   \item orthogonal reflection or symmetry;
   \item central symmetry;
   \item orthogonal projection;
   \item rotation (degrees);
   \item inversion.
\end{itemize}

The choice of transformations is made through the options. The macro is \tkzcname{tkzDefCircleBy} and the other for the transformation of a list of points \tkzcname{tkzDefCirclesBy}. For example, we'll write:
\begin{tkzltxexample}[]
\tkzDefCircleBy[translation= from A to A'](O,M) 
\end{tkzltxexample}
$O$ is the center and $M$ is a point on the circle.
The image is a circle. The new center is |tkzFirstPointResult| and |tkzSecondPointResult| is a point on the new circle. You can get the results with the macro \tkzcname{tkzGetPoints}.
\medskip
\begin{NewMacroBox}{tkzDefCircleBy}{\oarg{local options}\parg{pt1,pt2}}%
The argument is a couple of points. The results is a couple of points. If you want to keep these points then the macro \tkzcname{tkzGetPoints\{O'\}\{M'\}} allows you to assign the name \tkzname{O'} to the center and \tkzname{M'} to the point on the circle.

\begin{tabular}{lll}%
\toprule
arguments &amp;  definition &amp; examples               \\ 
\midrule
\TAline{pt1,pt2}   {existing points}   {$(O,M)$}
\bottomrule
\end{tabular}

\begin{tabular}{lll}%
options     &amp;     &amp; examples                         \\ 
\midrule
\TOline{translation}{= from \#1 to \#2}{[translation=from A to B](O,M)}
\TOline{homothety}  {= center \#1 ratio \#2}{[homothety=center A ratio .5](O,M)}
\TOline{reflection} {= over \#1--\#2}{[reflection=over A--B](O,M)}
\TOline{symmetry }  {= center \#1}{[symmetry=center A](O,M)}
\TOline{projection }{= onto \#1--\#2}{[projection=onto A--B](O,M)}
\TOline{rotation }  {= center \#1 angle \#2}{[rotation=center O angle 30](O,M)}
\TOline{inversion}{= center \#1 through \#2}{[inversion =center O through A](O,M)} 
% \TOline{inversion negative}{= center \#1 through \#2}{[inversion negative =center O through A](O,M)}
\bottomrule
\end{tabular}

\medskip
\emph{The image is only defined and not drawn.}
\end{NewMacroBox} 

\subsubsection{\tkzname{Translation}}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[&gt;=latex] 
 \tkzDefPoint(0,0){A}  \tkzDefPoint(3,1){B}
 \tkzDefPoint(3,2){C}   \tkzDefPoint(4,3){D}
 \tkzDefCircleBy[translation= from B to A](C,D) 
 \tkzGetPoints{C'}{D'} 
 \tkzDrawPoints[teal](A,B,C,D,C',D')
 \tkzDrawSegments[orange,-&gt;](A,B)
 \tkzDrawCircles(C,D C',D')
 \tkzLabelPoints[color=teal](A,B,C,C') 
 \tkzLabelPoints[color=teal,above](D,D') 
\end{tikzpicture} 
\end{tkzexample}

\subsubsection{\tkzname{Reflection} (orthogonal symmetry)}

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[&gt;=latex] 
 \tkzDefPoint(0,0){A}  \tkzDefPoint(3,1){B}
 \tkzDefPoint(3,2){C}   \tkzDefPoint(4,3){D}
 \tkzDefCircleBy[reflection = over A--B](C,D)
 \tkzGetPoints{C'}{D'} 
 \tkzDrawPoints[teal](A,B,C,D,C',D')
 \tkzDrawLine[add =0 and 1][orange](A,B)
 \tkzDrawCircles(C,D C',D')
 \tkzLabelPoints[color=teal](A,B,C,C') 
 \tkzLabelPoints[color=teal,right](D,D') 
\end{tikzpicture} 
\end{tkzexample}

\subsubsection{\tkzname{Homothety}}

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1.2]
 \tkzDefPoint(0,0){A}   \tkzDefPoint(3,1){B}
 \tkzDefPoint(3,2){C}   \tkzDefPoint(4,3){D}
 \tkzDefCircleBy[homothety=center A ratio .5](C,D)
 \tkzGetPoints{C'}{D'}
 \tkzDrawPoints[teal](A,C,D,C',D')
 \tkzDrawCircles(C,D C',D')
 \tkzLabelPoints[color=teal](A,C,C')
 \tkzLabelPoints[color=teal,right](D,D') 
\end{tikzpicture}
\end{tkzexample}

\subsubsection{\tkzname{Symmetry}}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1]
 \tkzDefPoint(3,1){B}
 \tkzDefPoint(3,2){C}   \tkzDefPoint(4,3){D}
 \tkzDefCircleBy[symmetry=center B](C,D)
 \tkzGetPoints{C'}{D'}
 \tkzDrawPoints[teal](B,C,D,C',D')
 \tkzDrawLines[orange](C,C' D,D')
 \tkzDrawCircles(C,D C',D')
 \tkzLabelPoints[color=teal](C,C')
 \tkzLabelPoints[color=teal,above](D)
 \tkzLabelPoints[color=teal,below](D')
\end{tikzpicture}
\end{tkzexample}

\subsubsection{\tkzname{Rotation}}
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=0.5]
 \tkzDefPoint(3,-1){B}
 \tkzDefPoint(3,2){C}   \tkzDefPoint(4,3){D}
 \tkzDefCircleBy[rotation=center B angle 90](C,D)
 \tkzGetPoints{C'}{D'}
 \tkzDrawPoints[teal](B,C,D,C',D')
 \tkzLabelPoints[color=teal](B,C,D,C',D')
 \tkzDrawCircles(C,D C',D')
\end{tikzpicture}
\end{tkzexample}

\subsubsection{\tkzname{Inversion}}

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1.5]
\tkzSetUpPoint[size=3,color=red,fill=red!20]
\tkzSetUpStyle[color=purple,ultra thin]{st1}
\tkzSetUpStyle[color=cyan,ultra thin]{st2}
\tkzDefPoint(2,0){A} \tkzDefPoint(3,0){B} 
\tkzDefPoint(3,2){C} \tkzDefPoint(4,2){D} 
\tkzDefCircleBy[inversion = center B through A](C,D) 
\tkzGetPoints{C'}{D'}
\tkzDrawPoints(A,B,C,D,C',D') 
\tkzLabelPoints(A,B,C,D,C',D')
\tkzDrawCircles(B,A)
\tkzDrawCircles[st1](C,D)
\tkzDrawCircles[st2](C',D')
\end{tikzpicture}
\end{tkzexample}

\endinput</pre></body></html>