Tracing a border formed by circles

Question

Suppose I have some code that draws a few circles. These circles overlap each other in such a way that they form one "group" -- there are no groups of circles completely separated from the rest. For instance:

\documentclass{article}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\draw (3,1) circle[radius=1];
\draw (3,2) circle[radius=1.5];
\draw (1,2) circle[radius=0.7];
\draw (0.9,1.2) circle[radius=1.2];
\draw (1,0) circle[radius=0.8];
\end{tikzpicture}
\end{document}

I would like to be able to draw only the "border" formed by these circles, without any of the other lines. It would be like drawing the border of the silhouette formed by all the circles. It would look something like this:

Now for that one, I just painted over the lines in Paint.NET with a brush - if you look closely you can see the imperfections. I would prefer to be able to do it in LaTeX directly. I understand that one option would be to manually calculate the position of the arcs and draw them in. However, it would be nice if there was a way for LaTeX to do the calculations for me and draw the border of the circles, without me having to solve a large amount of intersection equations. Is this possible?

Answer 1

I don't know tikz from a Polynesian mythological icon, so I am sure someone can do better.

I first draw the OP's original figure of black circles, and then redraw the same items as white filled circles of slightly smaller radius.

\documentclass{article}
\usepackage{tikz}
\newcommand\eps{.02}
\begin{document}
\begin{tikzpicture}
\draw (3,1) circle[radius=1];
\draw (3,2) circle[radius=1.5];
\draw (1,2) circle[radius=0.7];
\draw (0.9,1.2) circle[radius=1.2];
\draw (1,0) circle[radius=0.8];
\color{white}
\draw[fill=white] (3,1) circle[radius=1-\eps];
\draw[fill=white] (3,2) circle[radius=1.5-\eps];
\draw[fill=white] (1,2) circle[radius=0.7-\eps];
\draw[fill=white] (0.9,1.2) circle[radius=1.2-\eps];
\draw[fill=white] (1,0) circle[radius=0.8-\eps];
\end{tikzpicture}
\end{document}

I'll add that Ignasi provides good advice in the comments, to use

circle[radius=1cm-.5\pgflinewidth]

rather than my empirically derived value of \eps. In that way, the result will look good, regardless of any variations in preset line thickness.

Answer 2

Pretty much exactly the same approach as Steven B. Segletes but more (possibly overly) concise. The thickness of the outline is half the value specified in the line width:

\documentclass[tikz,border=5]{standalone}
\begin{document}
\tikz\path [draw, line width=5pt, postaction={fill=white}]
   \foreach \p/\r in {(3,1)/1, (3,2)/1.5, (1,2)/.7, (.9,1.2)/1.2, (1,0)/.8}{
      \p circle [radius=\r]};
\end{document}

And here is a outline fading decoration which makes the inside of the resulting shape transparent. It is mostly proof-of-concept, and as it uses fadings may not work with some viewers. Also it exploits the nonzero rule for filling so its successful use is dependent on the direction of the paths being drawn. The line width is set using the amplitude key.

\documentclass[tikz,border=5]{standalone}
\usetikzlibrary{fadings,decorations,fit}
\pgfdeclaredecoration{outline fading}{final}{
\state{final}{%
  \pgftransformreset%
  \pgfnodealias{@}{current path bounding box}%
  \node [fit=(@), inner sep=\pgfdecorationsegmentamplitude] (@@) {};
  \let\outlinepath=\pgfdecoratedpath
  \pgfinterruptpicture%
    \begin{tikzfadingfrompicture}[name=.]
      \path [line width=\pgfdecorationsegmentamplitude*2, draw=transparent!0,
       postaction={fill=transparent, nonzero rule}] 
       \pgfextra{\pgfsetpath\outlinepath};
      \useasboundingbox (@@.south west) (@@.north east);
    \end{tikzfadingfrompicture}%
  \endpgfinterruptpicture%
  \fill [path fading=.,fit fading=false,fading transform={shift={(@@)}}] 
   (@@.south west) rectangle (@@.north east); 
}}
\begin{document}
\begin{tikzpicture}
 \draw [help lines] (-1,-1) grid (5,5);
  \path [decoration={outline fading, amplitude=2pt}, decorate]
    \foreach \p/\r in {(3,1)/1, (3,2)/1.5, (1,2)/.7, (.9,1.2)/1.2, (1,0)/.8}{
        \p circle [radius=\r]};
\end{tikzpicture}
\end{document}

Answer 3

Here's a solution using the magic of the shadows library:

\documentclass[tikz, border=10pt]{standalone}
\usetikzlibrary{shadows}    
\begin{document}
  \begin{tikzpicture}[
        general shadow/.default={shadow scale=1, draw=black, line width=1pt, fill=none, shadow xshift=0pt, shadow yshift=0pt},
        my shadow/.append style={fill=white, general shadow},
      ]
    \path [my shadow]
      (3,1) circle [radius=1]
      (3,2) circle [radius=1.5]
      (1,2) circle [radius=0.7]
      (0.9,1.2) circle [radius=1.2]
      (1,0) circle [radius=0.8];
  \end{tikzpicture}
\end{document}