TikZ: Drawing an arc from an intersection to an intersection

Question

I want to draw an arc with a given radius from the point the arc intersects with a given circle to the point where the arc intersects with another circle. I have been reading through a lot of the TikZ manual, but to no avail. Can anyone help?

This PDF (http://www.mycroft.ch/tikztest.pdf) (MWE below) is intended to illustrate the problem. There are four circles, a part of each I want to use as path. The resulting curved wedge is then to be filled in black.

The green arc is no problem, I have start angle and end angle for it. For the two orange arcs I only have one angle (the one near the red circle is just an estimate). For the red arc I have nothing (both ends are estimates, thought to illustrate).

Interestingly, I can calculate the intersections (marked by the red circles), but I cannot fathom how to draw an arc there.

Hopefully I am overlooking something incredibly obvious! Thanks.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections,scopes}
\tikzset{
  pics/carc/.style args={#1:#2:#3}{
    code={
      \draw[pic actions] (#1:#3) arc(#1:#2:#3);
    }
  }
}
\begin{document}
\begin{tikzpicture}
\draw [name path=three] (120:1.06) circle (1.9);
\draw [name path=four] (0:1.06) circle (2.12);
\draw [name path=five] (0:0.77) circle (2.41);
\draw [name path=two] (0:0) circle (1.06);
\draw[green, thick] (0:3.18) arc [radius=2.12, start angle=0, end angle=180]; 
\draw[orange, thick] (0:3.18) arc [radius=2.41, start angle=0, end angle=197]; 
\draw[orange, thick] (180:1.06) arc [radius=1.06, start angle=180, end angle=245]; 
\draw[red, thick, name intersections={of=five and three}] (intersection-2) circle (2pt) node {}; 
\draw[red, thick, name intersections={of=two and three}] (intersection-1) circle (2pt) node {};
{ [xshift=-0.53cm,yshift=0.918cm] \pic [red,thick] {carc=238:274:1.9}; }
\end{tikzpicture}
\end{document}

Answer 1

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

1. 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.

2. 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}

Answer 2

This one is makes a nice example of using buildcycle with subpath in Metapost.

prologues := 3;
outputtemplate := "%j%c.eps";

beginfig(1);

path A, B, C, D, F;

A = fullcircle scaled 240;
B = fullcircle scaled 200 shifted 20 right;
C = fullcircle scaled 100 shifted 30 left;
D = fullcircle scaled 180 shifted 60 left shifted 40 up;

F = buildcycle(subpath (0,5) of A, 
               subpath (4,7) of D, 
               subpath (6,4) of C,
               subpath (4,0) of B);

fill F withcolor .8[blue,white];
draw A; draw B; draw C; draw D;
endfig;
end.

Each fullcircle has eight points numbered counter-clockwise from 3 o'clock. So subpath (0,5) of A is the an arc of the first 5/8 of A running counter-clockwise. If you reverse the order of the arguments to subpath you get a reversed path, so subpath (4,0) of B is the upper half of B running clockwise.

The buildcycle works best with a sequence of overlapping paths all running in the same direction.

Answer 3

I know that this is a very old question, but here's a solution using the new spath3 TikZ library that can work with splitting paths at intersection points. Using that, the various circles are split into sections which are then used to define the filled region and the coloured pieces around it.

(There's one extra path because TikZ wasn't picking up all the intersection points, especially where the circles are tangent.)

\documentclass{article}
%\url{https://tex.stackexchange.com/q/238967/86}
\usepackage{tikz}
\usetikzlibrary{intersections,scopes,spath3}
\begin{document}
\begin{tikzpicture}
% Draw and save the circles.
\draw [spath/save=three] (120:1.06) circle[radius=1.9];
\draw [spath/save=four] (0:1.06) circle[radius=2.12];
\draw [spath/save=five] (0:0.77) circle[radius=2.41];
\draw [spath/save=two] (0:0) circle[radius=1.06];
% This is useful for where the circles are meant to be tangent but don't actually quite intersect.
\path[overlay,spath/save=line] (-5,0) -- (5,0);
\tikzset{
  % Split various paths where they intersect with other paths:
  %
  % Split both three and two where they intersect
  spath/split at intersections={three}{two},
  % Split both three and five where they intersect
  spath/split at intersections={three}{five},
  % Split each of two, four, five where they intersect with line
  spath/split at intersections with={five}{line},
  spath/split at intersections with={four}{line},
  spath/split at intersections with={two}{line},
  % Get the components of each path as a list
  spath/get components of={three}\threeCpt,
  spath/get components of={five}\fiveCpt,
  spath/get components of={four}\fourCpt,
  spath/get components of={two}\twoCpt,
  % Join some of the components of path five together as it's been split into a lot of pieces
  spath/clone={five part}{\getComponentOf\fiveCpt{5}},
  spath/join with={five part}{\getComponentOf\fiveCpt{2},weld},
  spath/join with={five part}{\getComponentOf\fiveCpt{3},weld}
}
% Fill the region defined by the pieces
\fill[cyan,ultra thick,
  spath/use={\getComponentOf\fourCpt{2}},
  spath/use={\getComponentOf\twoCpt{5},weld},
  spath/use={\getComponentOf\threeCpt{5},weld,reverse},
  spath/use={five part,weld,reverse},
];
% Draw the pieces different colours
\draw[spath/use={\getComponentOf\threeCpt{5}},green,ultra thick];
\draw[spath/use={five part},red,ultra thick];
\draw[spath/use={\getComponentOf\fourCpt{2}},orange,ultra thick];
\draw[spath/use={\getComponentOf\twoCpt{5}},blue,ultra thick];
\end{tikzpicture}
\end{document}

Answer 4

\documentclass{standalone} 
\usepackage{tkz-euclide}

\begin{document} 
\begin{tikzpicture}
\tkzDefPoints{0/0/O1,2/0/I,-1/2/O2,-2/0/J,1/0/O3,6/0/K}
\tkzDefPoints{-2/-2/M,1/-2/N}
\tkzDrawCircles(O1,I O2,I I,J O3,K)
\begin{scope}
  \tkzClipCircle(O2,I)
  \tkzClipCircle[out](O1,I)
  \tkzClipCircle[out](I,J)
\tkzFillCircle[teal!20](O3,K)
\end{scope}

\begin{scope}
  \tkzClipCircle(O2,I)
  \tkzClipCircle[out](O1,I)
  \tkzClipCircle(I,J)
  \tkzClipPolygon(M,N,I,J)
\tkzFillCircle[teal!20](O3,K)
\end{scope}

\begin{scope}
  \tkzClipCircle[out](O2,I)
  \tkzClipCircle[out](I,J)
  \clip   (current bounding box.west) --(current bounding box.east)
     --(current bounding box.north east) --  (current bounding box.north west)
     -- cycle;
  \tkzFillCircle[teal!20](O3,K)
\end{scope}

\tkzDrawCircles(O1,I O2,I I,J O3,K)
\tkzDrawCircle(I,J)
\end{tikzpicture}
\end{document}

Answer 5

1. Circles with name path.
2. Finding intersections with name intersections.
3. Draw arcs between these points with arcto.

The points C2.0 = C3.0 and C2.180 = C4.180 are given explicitly because we know these points already (and I don't trust the intersections library with these touching points).

Code

\documentclass[tikz]{standalone}
\usetikzlibrary{
  intersections,
  through,        % → circle through
  ext.paths.arcto,% ← https://ctan.org/pkg/tikz-ext
}
\begin{document}
\begin{tikzpicture}[
  circle at/.style args={#1)#2radius#3}{
    at={#1)}, circle through={([shift={#1)}]0:#3)}, node contents=},
  declare function={x_radius=\pgfkeysvalueof{/tikz/x\space radius};},
  r1/.style={radius=1.9},  r2/.style={radius=2.12},
  r3/.style={radius=2.41}, r4/.style={radius=1.06},
  arc to/.search also=/tikz,
  %
  @/.code args={#1/#2}{
    \node(C#1)[draw,r#1,circle at={(#2) radius x_radius},name path=C#1];},
  @/.list={1/(120:1.06), 2/(0:1.06), 3/(0:0.77), 4/(0:0)},
  @/.style args={#1/#2}{name intersections={of=#1 and #2, name=#1/#2}},
  @/.list={C3/C1, C1/C4}
]
\fill[black]
  (C2.0) arcto[r3, large]     (C3/C1-2)
         arcto[r1]            (C1/C4-1)
         arcto[r4, clockwise] (C2.180)
         arcto[r2, clockwise] cycle;
\end{tikzpicture}
\end{document}

Output

Answer 6

I propose a construction that treats all the arcs on the same footing as arcs on certain circles. I defined arc on circle to which depends on two arguments, the next point on the circle, B, and the center of the circle, C. The output is the arc from the current point to B on the circle of center C. It needs the library math to handle the ifs in the insert path.

I tried to keep the notation from the question, this is why the index goes from 2 ot 5.

The code

\documentclass[margin=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{intersections}
\usetikzlibrary{math}
\tikzset{
  arc on circle to/.style args={#1, center=#2}{%  second point, center
    insert path={%
      coordinate (tmpPoint)
    },
    evaluate={%
      real \a, \b, \r;
      coordinate \C, \A, \B;
      \A = (tmpPoint);
      \B = (#1);
      \C = (#2);
      \r = veclen(\Ax-\Cx, \Ay-\Cy)*1pt/1cm;
      \a = atan2(\Ay-\Cy, \Ax-\Cx);
      \b = atan2(\By-\Cy, \Bx-\Cx);
      if \a<0 then {\a = \a +360;};
      if \b<0 then {\b = \b +360;};
    },
    insert path={%
      arc (\a: \b: \r)
    }
  }  
}
\begin{document}
\begin{tikzpicture}[evaluate={%
    real \r2, \r3, \r4, \r5;
    \r2 = 1.06;
    \r3 = 1.9;
    \r4 = 2.12;
    \r5 = 2.41;
  }]
  \path
  (0:0) coordinate (C2)
  (120:1.06) coordinate (C3)
  (0:1.06) coordinate (C4)
  (0:0.77) coordinate (C5);
\foreach \k in {2,...,5}{%
    \draw[name path=circle\k] (C\k) circle (\r\k);
  }
\path (C4) ++(0: \r4) coordinate (P54);
  \path[name intersections={of=circle4 and circle2, by={P42}}];
  \path[name intersections={of=circle2 and circle3, by={P23, Q23}}];
  \path[name intersections={of=circle3 and circle5, by={Q35, P35}}];
\draw[fill=blue!70, opacity=.2]
  (P54) [arc on circle to={P42, center=C4}]
  [arc on circle to={P23, center=C2}]
  [arc on circle to={P35, center=C3}]
  [arc on circle to={P54, center=C5}]
  -- cycle;
\end{tikzpicture}
\end{document}

Answer 7

I assume that I haven't been overlooking anything really obvious (because then someone would have pointed it out, probably), so I chose the less appealing, but more or less sufficient way to figure out the angles by hand:

\path [name path=A] (0:0.77) circle (2.41);
\path [name path=B, rotate=120] (0:1.06) circle (1.9);

\fill[black, name intersections={of=A and B}] (-1.06,0) arc [radius=2.12, start angle=180, end angle=0] arc [radius=2.41, start angle=0, end angle=196.75] arc [radius=1.9, start angle=238, end angle=273.8] arc [radius=1.06, start angle=247.54, end angle=180]  -- cycle; 

Thanks!