Tikz - how to draw a circle from one point to another?

Question

In a tikzpicture, I want to draw a circular arc that is almost all of a circle from one point to another. I have the coordinates of the start and end points and the radius I would like, but I do not know where to put the center of the circle. Naively, I expect that using draw with bend=N with N being slightly less than 360 would give the result I want, but when N exceeds 90, bend behaves as if a smaller number was given and the arc is more or less straight from the start point to the end point. However, given the plethora of features in Tikz, I expect that there is some straightforward way to get this effect.

In my particular application, the arc starts downward from the bottom of a node, circles around and ends downward into the top of the same node. My current best version is

\documentclass{amsart}

\usepackage{tikz}

\begin{document}

\begin{tikzpicture}
\node (box) at (0,-2) {$foo$};
\draw [->] (box.south)
      to [out=260,in=100,looseness=10]
      node [midway,left=0pt] {$bar$}
      (box.north);
\end{tikzpicture}

\end{document}

where "foo" is the text in the node and "bar" is a label on the arc. But it produces an elliptical arc rather than a circular arc:

Answer 1

If you just want to specify two points and a radius you can use the ext.paths.arcto library of my tikz-ext package.

Since there exist two circles with one radius through two points and thus four arcs, you will need to specify which arc one you want by using, in this case, the large and the clockwise flag.

Code

\documentclass{amsart}
\usepackage{tikz}
\usetikzlibrary{ext.paths.arcto}
\begin{document}
\begin{tikzpicture}
\node (box) at (0,-2) {$foo$};
\draw [->] (box.260) arc to[radius=1cm, large, clockwise]
  node [midway, left] {$bar$} (box.100);
\end{tikzpicture}
\end{document}

Output

Answer 2

Building on DraUX's answer, you can get Tikz to calculate start/end angles based on the radius. In this case:

\documentclass{amsart}

\usepackage{tikz}
\usetikzlibrary{calc}

\begin{document}

\begin{tikzpicture}
\node (box) at (0,-2) {$foo$};
\draw [->]
 let \n1={1cm},
 \p1=($(box.north)-(box.south)$),
 \n2={asin(\y1/(2*\n1))},
 \n3={360-\n2} in
 \pgfextra{\xdef\cstart{\n3}} \pgfextra{\xdef\cend{\n2}}
 (box.south) arc (\cstart:\cend:\n1) node [midway,left=0pt] {$bar$};
\end{tikzpicture}

\end{document}

Here, I have used \pgfextra to make the values of \n2 and \n3 visible globally (as \cstart and \cend). You can omit the \pgfextra's if you say "arc (\n3:\n2:\n1)".

Answer 3

You can achieve it by doing something like that

\documentclass{standalone}
\usepackage{standalone}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
    \node (box) at (0,-2) {$foo$};
    \draw (box.south) arc (160:-160:-0.7) node [midway,left=0pt] {$bar$};
\end{tikzpicture}
\end{document}

It works, but it is an ugly solve, since you have to change the start and end angles when you change the radius.

Answer 4

Like this:

with a minimal code (the line of grid is optional):

\documentclass[tikz,border=1cm]{standalone}

\begin{document}
    \begin{tikzpicture}
        \draw[gray!20] (-6,-6) grid (6,6);
        \draw[line width=3pt,latex-] (5:5) arc(5:355:5);
        \node at (0:5) {\Large foo};
        \node at (180:5.5) {\Large bar};
    \end{tikzpicture}
\end{document}