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This question already has an answer here:

I want to define the point X dynamically. Now I have defined the point at -90° but I want to make it changeable depending on the place of O. (I know there are similar questions, but I don't find the answer on my problem.)

enter image description here

 \documentclass{article}
 \usepackage{pgfplots}
 \usepackage{tkz-euclide}
 \usetkzobj{all}
 \begin{document}
 \begin{center}
\begin{tikzpicture}
\coordinate (M) at (0,0) ;
\coordinate (O) at (canvas polar cs:angle=90,radius=3cm) ;
\coordinate (A) at (canvas polar cs:angle=-130,radius=3cm);
\coordinate (B) at (canvas polar cs:angle=-30,radius=3cm) ;
\coordinate (X) at (canvas polar cs:angle=-90,radius=3cm);
\draw (M) circle (3cm);
\draw (A) -- (M) -- (B);
\draw (A) -- (O) -- (B);
\tkzDrawLine[dashed, add= 0.2 and 1.3,color=black](O,M)
\tkzDrawPoints(O,A,B,M,X);
\tkzLabelPoints[left](A,M);
\tkzLabelPoints[below right](X);
\tkzLabelPoints[above left](O);
\tkzLabelPoints(B);
\tkzMarkAngle[fill= red,size=1.5cm, opacity=.4](A,M,B);
\tkzMarkAngle[fill= red,size=1.5cm, opacity=.4](A,O,B);
\tkzLabelAngle[pos = 1.2](A,O,M){$2$};  
\tkzLabelAngle[pos = 1.2](M,O,B){$1$}; 
\tkzLabelAngle[pos = 1.1](A,M,X){$2$};  
\tkzLabelAngle[pos = 1.1](X,M,B){$1$};  
\end{tikzpicture}
\end{center}
\end{document}

marked as duplicate by jubobs, Qrrbrbirlbel, Thorsten, Heiko Oberdiek, egreg Aug 24 '13 at 16:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3

It's not clear from the question, how X shall be defined, but I guess you want the point opposite of O on the circle line. In that case, it's easiest, if you simply reflect O at M to get X:

 \documentclass{article}
 \usepackage{pgfplots}
 \usepackage{tkz-euclide}
 \usetkzobj{all}
 \begin{document}
 \begin{center}
\begin{tikzpicture}
\coordinate (M) at (0,0) ;
\coordinate (O) at (canvas polar cs:angle=110,radius=3cm) ;
\coordinate (A) at (canvas polar cs:angle=-130,radius=3cm);
\coordinate (B) at (canvas polar cs:angle=-30,radius=3cm) ;
% \coordinate (X) at (canvas polar cs:angle=-90,radius=3cm);
\tkzDefPointBy[symmetry=center M](O)\tkzGetPoint{X}
\draw (M) circle (3cm);
\draw (A) -- (M) -- (B);
\draw (A) -- (O) -- (B);
\tkzDrawLine[dashed, add= 0.2 and 1.3,color=black](O,M)
\tkzDrawPoints(O,A,B,M,X);
\tkzLabelPoints[left](A,M);
\tkzLabelPoints[below right](X);
\tkzLabelPoints[above left](O);
\tkzLabelPoints(B);
\tkzMarkAngle[fill= red,size=1.5cm, opacity=.4](A,M,B);
\tkzMarkAngle[fill= red,size=1.5cm, opacity=.4](A,O,B);
\tkzLabelAngle[pos = 1.2](A,O,M){$2$};  
\tkzLabelAngle[pos = 1.2](M,O,B){$1$}; 
\tkzLabelAngle[pos = 1.1](A,M,X){$2$};  
\tkzLabelAngle[pos = 1.1](X,M,B){$1$};  
\end{tikzpicture}
\end{center}
\end{document}

enter image description here

Of course, you might indeed want the second intersection of OM and the circle, which is a different point iff O is not on the circle line. In that case replace the definition of X by

\tkzInterLC(O,M)(M,A)\tkzGetPoints{notinteresting}{X}
  • 1
    TikZ solutions include \coordinate (X) at ([rotate around=180:(M)] O); and \coordinate (X) at ($(M)!-1!(O)$); – Qrrbrbirlbel Aug 24 '13 at 15:49
  • 1
    Or \coordinate (X) at ($2*(M)-(O)$);. – Toscho Aug 24 '13 at 15:55

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