Im not really adept at drawing with tikz, could anyone help me with a specific diagram? i need a circle or an ellipsoid with two curves on it that do not intersect.

Something like picture included

thank youenter image description here

  • 2
    Welcome to TeX.SX! Please add the code you have tried so far, so we can see what set-up you are working in. Nov 9, 2014 at 15:47
  • Your question leaves all the effort to our community, even typing the essentials of a TeX document such as \documentclass{}...\begin{document} etc. As it is, most of our users will be very reluctant to touch your question, and you are left to the mercy of our procrastination team who are very few in number and very picky about selecting questions. You can improve your question by adding a minimal working example (MWE) that more users can copy/paste onto their systems to work on. If no hero takes the challenge we might have to close your question.
    – cfr
    Nov 9, 2014 at 16:35

1 Answer 1


Not very tidy but you can polish it a bit:


  \begin{tikzpicture}[font=\sffamily, thick, outer sep=0pt]
    \node (a) [draw, ellipse, minimum width=45pt, minimum height=20pt] {A\hskip 15pt\ };
    \begin{scope}[on background layer]
      \path [draw=red, thick, rounded corners=5pt] (a.175) -- ($(a.130) + (0,17.5pt)$) coordinate (b) -- (a.north);
      \path [draw=blue, thick, rounded corners=5pt] (a.north) -- ($(a.east) + (5pt,20pt)$) coordinate (c) -- (a.east);
      \node [red] at ($(b)!2/3!(a.130)$) {B};
      \node [blue] at ($(c)!2/3!(a.25)$) {C};



  • Might be a good idea to set outer sep=0 on the \node, and perhaps redraw the node afterwards to get the red and blue lines to be underneath. Nov 9, 2014 at 17:19
  • @PeterGrill I initially put the red and blue underneath but it looked rather odd so I put them on top instead. Why zero outer sep?
    – cfr
    Nov 9, 2014 at 18:40
  • Setting outer sep=0 for the first \node yields better intersection of the lines. Nov 9, 2014 at 18:43
  • @PeterGrill Oh, I see. Yes, thank you. I've edited the answer.
    – cfr
    Nov 9, 2014 at 18:45

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