# How to draw arc in Tikz? [duplicate]

Possible Duplicate:
Best way to create this image, square and angle

How do I draw an arc between lines OA and OB to indicate an angle?

\begin{figure}
{
\begin{tikzpicture}[scale=2]
\node [draw, thick, minimum size=3.8cm, circle] at (0,2) {};
\node [draw, thick, minimum size=3cm, circle] at (1.3,2) {};
\draw (0,2) -- (1.3,2);
\draw (0,2) -- (0.8,2.56);
\draw (0,2) -- (0.8,1.44);
\draw (0,2) -- (0,2.95);
\draw (0-0.1,2.5) node{$a$};
\draw(1.3,2)--(1.770,2.6);
\draw (1.4,2.3) node{$r$};
\draw (0-0.1,2) node{$O$};
\draw (0.8-0.1,2.56) node{$A$};
\draw (0.8-0.1,1.44) node{$B$};
\end{tikzpicture}
}
\caption{}
\end{figure}

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## marked as duplicate by percusse, Thorsten, qubyte, Marco Daniel, Gonzalo MedinaMar 23 '12 at 16:13

– percusse Mar 23 '12 at 15:25
I would also suggest some other changes as well to the code: Use \coordinate (A) at (0.8,2.56); to define a coordinate then you can use (A) in the code instead of (0.8,2.56). This will make things easier to read. For instance: \draw (O) -- (A); were (O) is the coordinate for the origin ((0,2) in your case). You can add nodes as part of the line \draw (O) -- (Top) node [midway, left] {$a$} where (Top) is (0,2.95). Instead of defining new coordinates for nodes, use relative positioning to place them \node [above] at (A) {$A$}. – Peter Grill Mar 23 '12 at 16:35

\begin{tikzpicture}[scale=2]
\node [draw, thick, minimum size=3.8cm, circle] at (0,2) {};
\node [draw, thick, minimum size=3cm, circle] at (1.3,2) {};
\draw (0,2) -- (1.3,2);
\draw (0,2) -- (0.8,2.56);
\draw (0,2) -- (0.8,1.44);
\draw (0,2) -- (0,2.95);
\draw (0-0.1,2.5) node{$a$};
\draw(1.3,2)--(1.770,2.6);
\draw (1.4,2.3) node{$r$};
\draw (0-0.1,2) node{$O$};
\draw (0.8-0.1,2.56) node{$A$};
\draw (0.8-0.1,1.44) node{$B$};
% This is the angle: Take that part of circle which comes inside this rectangle
\path[clip]  (0,1.86) -- (0.4,1.86) -- (0.4,2.14) -- (0,2.14) -- cycle;
% The circle itself
\node [draw, blue,  thick, minimum size=1cm, circle] at (0,2) {};
\end{tikzpicture}


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Thank you very much. – user12290 Mar 23 '12 at 16:37