How to draw only part of a circle given points

Question

given three points $v_1, v_2, v_3$ and a circle with center $v_1$ such that $v_2, v_3$ lie on its boundary, I want to remove the part of the circle not between $v_2$ and $v_3$. I know how to do this if the corresponding angles are known (for example in the code below I want to get the part of the circle between the angles 0 and 90). But I do not know how to compute this for not rectangular settings.

\documentclass{article}
\usepackage{graphicx}
\usepackage{tikz}
\usetikzlibrary{calc}
\usetikzlibrary{arrows}
\usetikzlibrary{positioning}
\begin{document}
\begin{tikzpicture}
\path 
(0,0) coordinate (v1) node[below]{$v_1$}
(1,0) coordinate (v2) node[below]{$v_2$}
(0,1) coordinate (v3) node[below]{$v_3$};
\draw (v1)--(v2) (v1)--(v3);
\draw (v1) circle(1);
\foreach \p in {v1,v2,v3}
\fill (\p) circle(2pt);
\end{tikzpicture}
\end{document}

If you know the radius of the circle and polar angles, or know how to compute them from Cartesian coordinates, a mice solution is @Jake's answer to this question: https://tex.stackexchange.com/questions/123158/tikz-using-the-ellipse-command-with-a-start-and-end-angle-instead-of-an-arc — Jhor, May 12 '22 at 11:33
Hello, thank you very much for your comment. I think my huge problem is that I do not know how to obtain the right angles. Given those I could simply use the arc command, but what I want to do is, to say start at $v(2)$ and run along the circle until you meet $(v3)$. — samabu, May 12 '22 at 11:53
To compute te length and angles, see https://tex.stackexchange.com/questions/39293/coordinates-a-b-compute-b-a-and-angle-between-x-and-b-a — Jhor, May 12 '22 at 11:56
You are asking about angles for "not rectangular settings". Please update your example to reflect this, so we can see what you mean. — hpekristiansen, May 12 '22 at 12:20

Ignasi · Answer 1 · 2022-05-12T12:36:29.070

If you know the angles you can use an arc command:

\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\fill (0,0) coordinate[label=below:$v_1$] (v1) circle(2pt);
\fill (30:2cm) coordinate[label=right:$v_2$] (v2) circle(2pt);
\fill (95:2cm) coordinate[label=$v_3$] (v3) circle(2pt);
\draw (v3)--(v1)--(v2) arc[start angle=30, end angle=95, radius=2];
\end{tikzpicture}
\end{document}

if you don't know the angles, calc library can help:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\foreach \i in {1,2,3}{
\begin{tikzpicture}
\fill (0,0) coordinate[label=below:$v_1$] (v1) circle(2pt);
\fill (360rand:2cm) coordinate[label=right:$v_2$] (v2) circle(2pt);
\fill (360rand:2cm) coordinate[label=$v_3$] (v3) circle(2pt);
\draw let \p1=($(v2)-(v1)$), \p2=($(v3)-(v1)$), \n1={veclen(\x1,\y1)}, \n2={atan2(\y1,\x1)},
\n3={atan2(\y2,\x2)} in (v3)--(v1)--(v2) arc[start angle=\n2, end angle=\n3, radius=\n1];
\end{tikzpicture}}
\end{document}

Hello, thank you for the answer. I am familiar with the arc command, but I do not know how to get the right angles. Do you know how to elegantly compute them? — samabu, May 12 '22 at 11:50
@samabu i don't know about its elegancy, but I've added a possible solution to my answer. — Ignasi, May 12 '22 at 12:37

score 0 · Answer 2 · answered May 12 '22 at 12:29

In your particular case the transform from cartesian coordinate to polar is very simple. By elementary geometry we know, that:

(0,1) --> (90:1)
(1,0) --> (0:1)

considering above the arc command is simple to define. So, your MWE can be changed to:

\documentclass[margin=3mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{arrows.meta,
                calc,
                positioning}
\begin{document}
    \begin{tikzpicture}[
dot/.style = {circle, fill, inner sep=1pt,
              label=#1, node contents={}},
every label/.append style = {inner sep=2pt, font=\footnotesize}
                        ] 
\draw   (0,0) node[dot=below:$v_1$] -- 
        (0,1) node[dot=above:$v_2$] arc(90:0:1)
              node[dot=below:$v_3$] -- cycle;
    \end{tikzpicture}
\end{document}

At more general cartesian coordinate, you need first to calculate star t and the end angle using inverse trigonometric function. For example:

\alpha = acos{\frac{x}{x^2 + y^2}

where x and y are cartesian coordinates of a point on the $x,y$ plane.

John Kormylo · Answer 3 · 2022-05-12T13:30:34.307

You can do it with \clip, although it is a lot easier for short arcs (blue) than long arcs (red).

\documentclass{standalone}
\usepackage{graphicx}
\usepackage{tikz}
\usetikzlibrary{calc}
\usetikzlibrary{arrows}
\usetikzlibrary{positioning}
\begin{document}
\begin{tikzpicture}
\path 
(0,0) coordinate (v1) node[below]{$v_1$}
(1,0) coordinate (v2) node[below]{$v_2$}
(0,1) coordinate (v3) node[below]{$v_3$};
\draw (v1)--(v2) (v1)--(v3);
\begin{scope}
\clip (v2) rectangle (v3);
\draw[blue] (v1) circle(1);
\end{scope}
\begin{scope}[local bounding box=Bob]
\path (v1) circle(1);
\clip (Bob.north west)--(v3)--(v1)--(v2)--(Bob.south east)--
  (Bob.south west)--cycle;
\draw[red] (v1) circle(1);
\end{scope}
\foreach \p in {v1,v2,v3}
\fill (\p) circle(2pt);
\end{tikzpicture}
\end{document}

How to draw only part of a circle given points

3 Answers3