TikZ: Drawing an arc between two coordinates?

Question

I am up with the following:

Using the code given at the bottom.

Suffice to say, I don't really like the result: I wanted to draw this picture without explicitly calculating the circle and the lines going into and out of it. However, I just don't know how to draw that part-circle to the right (with $\theta$ in it), such that I...

• get an arc which is shaped like a circle,
• have it intersect the shown without actually crossing them, and
• have $\theta$ placed between the middle of the actually drawn arc and its center.

How could I achieve this?

\begin{tikzpicture}

\coordinate (a) at (-3,0.5);
\coordinate (b) at (-1,0.5);

\draw[solid,-latex] (a) -- (b) coordinate[at start] (p);

\draw[thin,solid,latex-latex] (p) -- node[left] {$p$} ++(270:0.5);

\draw[thin,dashed] (-3,0) -- (3,0) coordinate[very near end] (angleStart);

\node (interaction) [circle through=(b)
                    , solid
                    , draw
                    , fill=lightgray
                    ]  at (0,0) {interaction};

\draw[solid,-latex] (interaction.north east)  -- ++(45:2) coordinate[near end] (angleEnd);

\draw (angleStart) to[out=50,in=-20] node[below left=2pt] {$\theta$} (angleEnd);
\end{tikzpicture}

Answer 1

I suggest that you base everything on the center of the circle in which case the calculations can easily be automated with the additional TikZ library: \usetikzlibrary{calc}

You can set the angle and radius you want via:

\newcommand*{\ArcAngle}{60}%
\newcommand*{\ArcRadius}{2.0}%

Everything else is computed based on these. You can adjust these to get the specific result you desire.

Two other tweaks that I have defined below are to adjust how far the dashed lines extend past the arc, and how far past the arc the label is placed (as percentages):

\newcommand*{\LineExtend}{1.25}%
\newcommand*{\LableExtend}{1.10}%

Code:

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}

\newcommand*{\ArcAngle}{60}%
\newcommand*{\ArcRadius}{2.0}%
\newcommand*{\LineExtend}{1.25}%
\newcommand*{\LableExtend}{1.10}%

\begin{document}
\begin{tikzpicture}
\pgfmathsetmacro{\XValueArc}{\ArcRadius*cos(\ArcAngle)}%
\pgfmathsetmacro{\YValueArc}{\ArcRadius*sin(\ArcAngle)}%

\pgfmathsetmacro{\XValueLabel}{\ArcRadius*cos(\ArcAngle/2)}%
\pgfmathsetmacro{\YValueLabel}{\ArcRadius*sin(\ArcAngle/2)}%

\coordinate (Origin) at (0,0);

\draw [thin, dashed] (Origin) -- ($(\LineExtend*\ArcRadius,0)$);% Horizontal

% Extend this past the (\XValue,\YValue)
\draw [thin, dashed] (Origin) -- ($\LineExtend*(\XValueArc,\YValueArc)$);

\node (interaction) [circle, solid, draw, fill=lightgray]  
    at (Origin) {interaction};

\draw [<->] ($(Origin)+(\ArcRadius,0)$) 
    arc (0:\ArcAngle:\ArcRadius);

\node at ($\LableExtend*(\XValueLabel,\YValueLabel)$) {$\theta$};
\end{tikzpicture}
\end{document}

Answer 2

With pgf 2.1 cvs for the node \theta

\documentclass{standalone}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}

\coordinate (Origin) at (0,0);
\node (interaction) [circle, draw, 
                     fill = lightgray] at (Origin) {interaction};    

\draw [thin, dashed] (interaction.0) -- ++(0:1.5) 
                     (interaction.60)   -- ++(60:1.5) ;

\draw [<->] (0:2.1)  arc (0:60:2.1) node [right,pos=0.5] {$\theta$};
\end{tikzpicture}
\end{document} 

Answer 3

Just for fun.

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-eucl,pstricks-add}

\begin{document}
\begin{pspicture}[showgrid=false](-3,-1)(3,3)
    \psset{PointName=none,PointSymbol=none}
    \pstGeonode
        (-2.5,+0.0){A}
        (-2.5,+0.5){B}
        (-0.5,+0.5){C}
        (+0.0,+0.0){D}
        (+2.0,+2.0){E}
        (+1.5,+0.0){F}
    \pstLineAB[nodesepB=-1,linestyle=dashed]{A}{F}
    \pstLineAB[arrows=->]{B}{C}
    \pstLineAB[linestyle=dashed,arrows=->]{D}{E}
    \pstMarkAngle[LabelSep=1.15,MarkAngleRadius=1.5,arrows=->]{F}{D}{E}{$\theta$}
    \pstCircleOA[fillstyle=solid,fillcolor=gray]{D}{C}
    \rput(D){\tiny Interaction}
    \ncline[offset=2pt]{<->}{A}{B}\naput{$p$}
\end{pspicture}
\end{document}

Answer 4

The easiest way to draw an arc from the center is

\draw (\CENTERx,\CENTERy) ++( 45 : 1 ) arc ( 45:0:1 );

Draws an arc on a circle of radius 1 centered at (\CENTERx,\CENTERy), from (45 degrees) to (0 degrees).

See here for an example