Can PSTricks or others draw the 4 common tangent lines of 2 “disjoint” circles without having to do extra calculations?

Question

There are two disjoint circles. Their centers and radii are given. Without doing extra calculations, can we draw the 4 tangent lines using PSTricks (preferred) or others?

I asked many Illustrator, Free-hand, CAD experts, they cannot do it exactly. :-)

Answer 1

pstricks-add knows a macro for calculating and saving 10 points as nodes, the two central points and the four points on the circle.

\documentclass{article}
\usepackage{pstricks-add}% loads pst-node

\begin{document}
\begin{pspicture}[showgrid](0,0)(10,10)
\pscircle(1,1){1}\pscircle(7,7){3}
\psCircleTangents(1,1){1}(7,7){3}
\pcline[nodesep=-1cm,linecolor=blue](CircleTO1)(CircleTO2)
\pcline[nodesep=-1cm,linecolor=blue](CircleTO3)(CircleTO4)
\pcline[nodesep=-1cm,linecolor=red](CircleTI1)(CircleTI2)
\pcline[nodesep=-1cm,linecolor=red](CircleTI3)(CircleTI4)
\psdots(CircleTC1)\psdots(CircleTC2)%
  (CircleTO1)(CircleTO2)(CircleTO3)(CircleTO4)%
  (CircleTI1)(CircleTI2)(CircleTI3)(CircleTI4)%
\end{pspicture}

\end{document}

the output looks like!

Answer 2

TikZ can work out the tangent line between a circle and a point, so it's halfway there for this. With a tiny bit of mathematics, this can be bootstrapped to the tangent lines you want. The following code will do it (though it ought to check for the case where the two radii are the same - at the moment, that will produce an error, as will the situation where the circles overlap).

Here's the result:

Here's the code:

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

\begin{document}
\begin{tikzpicture}
\pgfmathsetmacro{\rone}{3}
\pgfmathsetmacro{\rtwo}{2}
\pgfmathsetmacro{\mid}{\rone/(\rone + \rtwo)}
\pgfmathsetmacro{\out}{\rone/(\rone - \rtwo)}
\node[draw,circle,minimum size=2 * \rone cm,inner sep=0pt] (c1) at (1,0) {};
\node[draw,circle,minimum size=2 * \rtwo cm,inner sep=0pt] (c2) at (-1,-6) {};
\path (c1.center) -- node[coordinate,pos=\mid] (mid) {} (c2.center);
\path (c1.center) -- node[coordinate,pos=\out] (out) {}  (c2.center);

%\draw[red] (tangent cs:node=c2,point={(mid)}) -- (tangent cs:node=c1,point={(mid)});
%\draw[red] (tangent cs:node=c2,point={(mid)},solution=2) -- (tangent cs:node=c1,point={(mid)},solution=2);

%\draw[red] (tangent cs:node=c2,point={(out)}) -- (tangent cs:node=c1,point={(out)});
%\draw[red] (tangent cs:node=c2,point={(out)},solution=2) -- (tangent cs:node=c1,point={(out)},solution=2);

\foreach \i in {1,2}
\foreach \j in {1,2}
\foreach \k in {mid,out}
\coordinate (t\i\j\k) at (tangent cs:node=c\i,point={(\k)},solution=\j);

\foreach \i in {1,2}
\foreach \k in {mid,out}
\draw[red] ($(t1\i\k)!-1cm!(t2\i\k)$) --  ($(t2\i\k)!-1cm!(t1\i\k)$);

\end{tikzpicture}
\end{document}

The commented-out lines will draw the tangent lines to the exact points, I chose to extend them a little to show that they were genuinely tangent and that's what the code after the commented-out lines are for.

Despite being a mathematician, I didn't actually compute the formulae for the crossing points - I just guessed something that "felt right" and then tested it and it seems to work. However, I can't guarantee that it is right.

Answer 3

The next version of tkz-euclide can draw the tangent lines. I created two macro for the internal similitude center of two circles and the external similitude center.

\documentclass{scrartcl}
\usepackage[usenames,dvipsnames]{xcolor}  
\usepackage{tkz-euclide} 
\usetkzobj{all} 
\definecolor{fondpaille}{cmyk}{0,0,0.1,0}
\pagecolor{fondpaille}
\color{Maroon}   

\begin{document}  
\begin{tikzpicture}
   \tkzInit[xmin=-5,ymin=-5,xmax=5,ymax=5]
   \tkzDefPoint(0,0){O}
   \tkzDefPoint(4,-5){A}
   \tkzDrawCircle[R](O,3 cm)
   \tkzDrawCircle[R](A,2 cm) 
   \tkzIntSimilitudeCenter(O,3)(A,2) \tkzGetPoint{I}
   \tkzDrawPoint(I) 
   \tkzExtSimilitudeCenter(O,3)(A,2) \tkzGetPoint{J}
   \tkzDrawPoint(J) 
   \tkzTangent[from with R= I](O,3 cm)  \tkzGetPoints{D}{E} 
   \tkzTangent[from with R= I](A,2 cm)  \tkzGetPoints{D'}{E'}
   \tkzTangent[from  with R= J](O,3 cm) \tkzGetPoints{F}{G}
   \tkzTangent[from with R= J](A,2cm)   \tkzGetPoints{F'}{G'} 
   \tkzDrawSegments[color=red](I,D I,E I,D' I,E')   
   \tkzDrawSegments[color=blue](J,F J,G)
  \end{tikzpicture}     

\end{document}

the code is very simple to get these centers:

%<--------------------------------------------------------------------------–> 
%                    Internal Similitude center
%<--------------------------------------------------------------------------–>
\def\tkzIntSimilitudeCenter(#1,#2)(#3,#4){%
\begingroup
\path[coordinate]  (barycentric cs:#1=#4,#3=#2) coordinate (tkzPointResult);
\endgroup
}
%<--------------------------------------------------------------------------–> 
%                    External Similitude center
%<--------------------------------------------------------------------------–>
\def\tkzExtSimilitudeCenter(#1,#2)(#3,#4){%
\begingroup
 \path[coordinate]  (barycentric cs:#1=-#4,#3=#2) coordinate (tkzPointResult);
\endgroup
}

Then we can get the tangents. The code will be upload in a few days on the ctan servers.