4

Previously I posted a question and one answer brought me another question.

Here is the code:

\documentclass[margin=10pt]{standalone}
\usepackage[utf8]{inputenc}
\usepackage{tikz,tkz-euclide,xcolor,graphicx}
\usetkzobj{all}

\begin{document}
\begin{tikzpicture}

\tkzDefPoints{0/0/A, 5/0/B, 3.7896542/3/P}

\draw[thick](A)--(B);

\tkzDrawPoints[color=red](A,B)
\tkzLabelPoint[below=2pt](A){$A$}
\tkzLabelPoint[below=2pt](B){$B$}

\tkzDrawPoint[fill=yellow](P)
\tkzLabelPoint[above=2pt](P){$P$}

\tkzDrawAltitude(A,B)(P) \tkzGetPoint{p}
\tkzDrawSegment(P,p)

\tkzLabelPoint[below=2pt](p){$p$}

\end{tikzpicture}

\end{document} 

Which yields:

per.

Now my question. I want to draw a circle which has got the same radius as the length of the segment (P,p) and the middle of the circle is the point P itself. Is it possible? If yes, how to draw that?

  • I don't understand the question: If you know the distance P to p and the middle of the circle (P) -- isn't this all you need to know? What's A and B? – user31729 Dec 13 '15 at 19:00
  • @ChristianHupfer I suppose A and B were irrelevant in this case, but they were in his previous question. :D – Alenanno Dec 14 '15 at 8:28
  • @Alenanno: Ah, a X-Y-question – user31729 Dec 14 '15 at 10:20
4

This is described in chapter 15 of the documentation, "Les Cercles": the command for drawing a circle around P with a radius equal to the distance between P and p is \tkzDrawCircle(P,p):

\documentclass[margin=10pt]{standalone}
\usepackage[utf8]{inputenc}
\usepackage{tikz,tkz-euclide,xcolor,graphicx}
\usetkzobj{all}

\begin{document}
\begin{tikzpicture}

\tkzDefPoints{0/0/A, 5/0/B, 3.7896542/3/P}

\draw[thick](A)--(B);

\tkzDrawPoints[color=red](A,B)
\tkzLabelPoint[below=2pt](A){$A$}
\tkzLabelPoint[below=2pt](B){$B$}

\tkzDrawPoint[fill=yellow](P)
\tkzLabelPoint[above=2pt](P){$P$}

\tkzDrawAltitude(A,B)(P) \tkzGetPoint{p}
\tkzDrawSegment(P,p)

\tkzLabelPoint[below=2pt](p){$p$}

\tkzDrawCircle(P,p)

\end{tikzpicture}

\end{document} 
  • 2
    +1 Argh, I was about to answer! :D But your solution is easier than mine, so that's for the better. – Alenanno Dec 13 '15 at 19:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.