A solution which allows to draw intersection segments of any two intersections is available as tikz library fillbetween.
This library works as general purpose tikz library, but it is shipped with pgfplots and you need to load pgfplots in order to make it work:
\documentclass{standalone}
\usepackage{tikz}
\usepackage{pgfplots}
\usetikzlibrary{fillbetween}
\begin{document}
\begin{tikzpicture}
\draw [name path=red,red] (120:1.06) circle (1.9);
%\draw [name path=yellow,yellow] (0:1.06) circle (2.12);
\draw [name path=green,green!50!black] (0:0.77) circle (2.41);
\draw [name path=blue,blue] (0:0) circle (1.06);
% substitute this temp path by `\path` to make it invisible:
\draw[name path=temp1, intersection segments={of=red and blue,sequence=L1}];
\draw[red,-stealth,ultra thick, intersection segments={of=temp1 and green,sequence=L3}];
\end{tikzpicture}
\end{document}
The key intersection segments is described in all detail in the pgfplots reference manual section "5.6.6 Intersection Segment Recombination"; the key idea in this case is to
create a temporary path temp1 which is the first intersection segment of red and blue, more precisely, it is the first intersection segment in the Left argument in red and blue : red. This path is drawn as thin black path. Substitute its \draw statement by \path to make it invisible.
Compute the desired intersection segment by intersecting temp1 and green and use the correct intersection segment. By trial and error I figured that it is the third segment of path temp1 which is written as L3 (L = left argument in temp1 and green and 3 means third segment of that path).
The argument involves some trial and error because fillbetween is unaware of the fact that end and startpoint are connected -- and we as end users do not see start and end point.
Note that you can connect these path segments with other paths. If such an intersection segment should be the continuation of another path, use -- as before the first argument in sequence. This allows to fill paths segments:
\documentclass{standalone}
\usepackage{tikz}
\usepackage{pgfplots}
\usetikzlibrary{fillbetween}
\begin{document}
\begin{tikzpicture}
\draw [name path=red,red] (120:1.06) circle (1.9);
%\draw [name path=yellow,yellow] (0:1.06) circle (2.12);
\draw [name path=green,green!50!black] (0:0.77) circle (2.41);
\draw [name path=blue,blue] (0:0) circle (1.06);
% substitute this temp path by `\path` to make it invisible:
\draw[name path=temp1, intersection segments={of=red and blue,sequence=L1}];
\draw[red,fill=blue,-stealth,ultra thick, intersection segments={of=temp1 and green,sequence=L3}]
[intersection segments={of=temp1 and green, sequence={--R2}}]
;
\end{tikzpicture}
\end{document}
\draw (20,12) -- ++(2,0) -- ++(0,2) -- ++(-3,0) -- ++(45:3);
Use ++ before each new incremental coordinate to make it relative to the last one and put the pencil there.
Here's a complete example:
\documentclass{article}
\usepackage{tikz}
\begin{document}
\tikz\draw (20,12) -- ++(2,0) -- ++(0,2) -- ++(-3,0) -- ++(30:3) {[rounded corners=10pt]-- ++(5,0) -- ++(0,-6)} -- ++(-7,0) -- cycle;
\end{document}
Of course, combining this with the -| or |- path operators can simplify the code even further; the following two pieces of code produce the same result:
\tikz\draw (20,12) -- ++(2,0) -- ++(0,2) -- ++(3,0) -- ++(0,1) -- ++(1,0) -- ++(0,-3) -- ++(2,0);\par\bigskip
and
\tikz\draw (20,12) -| ++(2,2) -| ++(3,1) -- ++(1,0) |- ++(2,-3);
I don't think that defining commands in this case adds anything; in fact, I think it reduces the functionality of the existing syntax (which is already simple). The example demonstrates that you can use, for example, polar coordinates and modify (up to TikZ limitations) the path attributes midways; even if the current question doesn't require this, it's a good thing to have the possibility to do those kind of modification if they are required.
Best Answer
The problem is that an
arcof radius 1 from one coordinate (call it A) which starts at angle X and ends at angle Y will assume that A is at angle of X on the circumference of a circle with radius 1.So your first
arcassumes that(0,0)is at zero degrees on the circumference of a circle with radius 1. This means the centre of the circle must be at(-1,0).But your second
arcassumes that(1,0)is at 120 degrees on the circumference of a circle with radius 1. So the centre must be at(1.5,{.5*(sqrt(3))}.Finally, your third
arcassumes that(0,0)is at 240 degrees on the circumference of a circle with radius 1. So the centre must be at({cos(60)},{sin(60)}).[On the improbable assumption that I've not messed up the geometry at this hour of the night.]
I suggest using polar coordinates instead. Then your
arcs can start at(0:1),(120:1)and(240:1)which makes things much simpler.