Cfrgtkky

documentclass[border=15pt,pstricks,12pt]{standalone}

usepackage{pst-eucl}

begin{document}

begin{pspicture}[showgrid](-5,-5)(10,10)

psset{unit=2cm,PointSymbol=none,PointName=none}

pstTriangle(0,4){C}(0,0){A}(3,0){B} 

pstMiddleAB{A}{B}{I1} pstArcOAB{I1}{A}{B}

pstMiddleAB{B}{C}{I2} pstArcOAB{I2}{B}{C}

pstMiddleAB{C}{A}{I3} pstArcOAB{I3}{C}{A}



end{pspicture}

end{document}

See the result from Asymptote(TranLeNam's code)

documentclass[border=5pt,varwidth]{standalone}

usepackage{asymptote}

begin{document}

begin{asy}

settings.outformat="pdf"; 

settings.prc=false;

settings.render=20;



unitsize(0.5cm);

import geometry;

defaultpen(fontsize(12pt));

pair A=(0,0), B=(8,0), C=(0,6);

draw(triangle(A,B,C));

pair CP=midpoint(A--B), BP=midpoint(A--C), AP=midpoint(C--B);

draw(arc(CP,A,B)^^arc(BP,C,A)^^arc(AP,B,C)) ;

inversion inv=inversion(20,A);

point AB=inv*B; 

point AC=inv*C; 

point AA=AB+AC; 

circle cABC=excircle(AB,AC,AA);

draw(inv*cABC,red);

shipout(bbox(1mm,1mm+white));

end{asy}

end{document}

Of course it does not have center and radius!! It uses inversion which PSTricks does not have !?

I truly do not know how to draw it for PSTricks.

More detail, see http://mathworld.wolfram.com/Inversion.html and http://www.piprime.fr/files/asymptote/geometry/modules/geometry.asy.html#struct%20inversion

Supposing that: A(0,0), B(b,0), C(0,c)

So AB is horizontal and AC is vertical we can calculate it. @jfbu did these calculations and I just implemented them into a PSTricks picture.

documentclass[pstricks,dvipsnames]{standalone}

usepackage{pst-eucl}



pagestyle{empty}



begin{document}



begin{pspicture}(-3,-3.5)(4.5,5.5)

pstVerb{%

%% Enter the coordinates of the points of the rectangled triangle

%% A(0,0), B(3,0), C(0,4) 

%% So AB is horizontal

%% AC is vertical

/xA 0 def

/yA 0 def

/xB 3 def

/yB 0 def

/xC 0 def

/yC 4 def

%% Midpoints of the sides

/mAx xB xC add 2 div def

/mAy yB yC add 2 div def

/mBx xA xC add 2 div def

/mBy yA yC add 2 div def

/mCx xA xB add 2 div def

/mCy yA yB add 2 div def

%% Calculating the radius of the half circles along the sides of the

%% rectangled triangle

/rAB xA xB sub 2 exp yA yB sub 2 exp add sqrt 2 div def

/AB rAB 2 mul def

/rAC xA xC sub 2 exp yA yC sub 2 exp add sqrt 2 div def

/AC rAC 2 mul def

/rBC xB xC sub 2 exp yB yC sub 2 exp add sqrt 2 div def

/BC rBC 2 mul def

/DENOM BC 5 mul AB 3 mul add AC 3 mul add def

%% @JFBU formula

%% X = 2 AB (AB + BC) / (5 BC + 3 AB + 3 AC)

%% Y = 2 AC (AC + BC) / (5 BC + 3 AB + 3 AC)

%% R = 2 (BC + AB) (BC + AC) /  (5 BC + 3 AB + 3 AC)

/x0 AB BC add AB mul 2 mul DENOM div def

/y0 AC BC add AC mul 2 mul DENOM div def

/r0 BC AB add BC AC add mul 2 mul DENOM div def

%% @JFBU formula for the tangent points

%% P1 = (2(a+c)c, -(a+b+c)c) / (3a + b + 2c)

%% P2 = (-(a+b+c)b, 2(a+b)b) / (3a + 2b + c)

%% P3 = ((a+b+c)(a+c), (a+b+c)(a+b))/ (3a + 2b + 2c)

/p1x BC AB add AB mul 2 mul BC 3 mul AC add AB 2 mul add div def

/p1y BC AC add AB add AB mul BC 3 mul AC add AB 2 mul add div neg def

/p2x BC AC add AB add AC mul BC 3 mul AC 2 mul add AB add div neg def

/p2y BC AC add AC mul 2 mul BC 3 mul AC 2 mul add AB add div def

/p3x BC AC add AB add BC AB add mul BC 3 mul AC 2 mul add AB 2 mul add div def

/p3y BC AC add AB add BC AC add mul BC 3 mul AC 2 mul add AB 2 mul add div def

}

%% Setting the nodes of the points of the triangle

pstGeonode[PointSymbol=none,PosAngle={225,-45,90}](!xA yA){A}(!xB yB){B}(!xC yC){C}

pspolygon[linecolor=blue,linejoin=1](A)(B)(C)

%% Setting the nodes of the midpoints of the triangle sides

pstMiddleAB[PointSymbol=none,PointName=none]{A}{B}{MAB}

pstMiddleAB[PointSymbol=none,PointName=none]{A}{C}{MAC}

pstMiddleAB[PointSymbol=none,PointName=none]{B}{C}{MBC}

%% Full circles along the sides of the triangle (not needed!)

%pscircle(MAB){!rAB}

%pscircle(MAC){!rAC}

%pscircle(MBC){!rBC}

%% Drawing the half circles along the sides of the triangle

pstArcOAB[linecolor=lightgray]{MAB}{A}{B}

pstArcOAB[linecolor=lightgray]{MAC}{C}{A}

pstArcOAB[linecolor=lightgray]{MBC}{B}{C}

%% circle and tangent points

pnode(!x0 y0){O}

pnode(!p1x p1y){P1}

pnode(!p2x p2y){P2}

pnode(!p3x p3y){P3}

pscircle[linecolor=Green](O){!r0}

psdot[linecolor=red](P1)uput[-90](P1){$P_1$}

psdot[linecolor=red](P2)uput[180](P2){$P_2$}

psdot[linecolor=red](P3)uput[90](P3){$P_3$}

pspolygon[linejoin=1,linecolor=red,linewidth=0.5pt](P1)(P2)(P3)

end{pspicture}

end{document}

Hope this answers the question.

As an animated gif:

For a triangle with a right angle:

documentclass[border=15pt,pstricks,12pt]{standalone}

usepackage{pst-eucl}

defpstInvCircle#1#2#3#4{%

  pnode(!

    psGetNodeCenter{#1}

    psGetNodeCenter{#2}psGetNodeCenter{#3} 

    #2.x #1.x sub #2.y #1.y sub Pyth /l12 ED

    #2.x #1.x sub #2.y #1.y sub #3.x #1.x sub #3.y #1.y sub Pyth2 /l23 ED

    #3.x #1.x sub #3.y #1.y sub Pyth /l31 ED 

    l23 5 mul l12 3 mul add l31 3 mul add /Denom ED

    l12 l23 add l12 mul 2 mul Denom div /xM ED

    l31 l23 add l31 mul 2 mul Denom div /yM ED

    l23 l12 add l23 l31 add mul 2 mul Denom div /rM ED

    xM yM ){#4}%

  pscircle[linecolor=red,linewidth=2pt,dimen=inner](#4){! rM }%

}

begin{document}

psset{unit=2}

begin{pspicture}[showgrid](-2,-2)(4,5)

psset{PointSymbol=none,PointName=none,dimen=inner,opacity=0.5}

pstTriangle[linecolor=blue,linewidth=2pt,linejoin=2](0,0){A}(3,0){B}(0,4){C} 

pstMiddleAB{A}{B}{I1} pstArcOAB[fillstyle=solid,fillcolor=red!40]{I1}{A}{B}

pstMiddleAB{B}{C}{I2} pstArcOAB[fillstyle=solid,fillcolor=blue!40]{I2}{B}{C}

pstMiddleAB{C}{A}{I3} pstArcOAB[fillstyle=solid,fillcolor=green!40]{I3}{C}{A}

psset{linecolor=red,linewidth=2pt,dimen=inner}

pstInvCircle{A}{B}{C}{O}

psdot(O)

end{pspicture}

end{document}

One possible way/starting point (but not accurate here) is to use something like this:

%&pdflatex

% !TeX TXS-program:compile = txs:///pdflatex/[--shell-escape]

documentclass[a4paper, pdf, x11names]{standalone}

usepackage{pstricks}

usepackage{pstricks-add, auto-pst-pdf}



usepackage{pst-eucl}

begin{document}

begin{pspicture}(-5,-5)(5,5)

pstTriangle[PointSymbol=none, PointName={A,B,A}](4,1){A}(1,3){B}(5,5){C}

pstCircleABC[CodeFig=true, CodeFigColor=white,linecolor=red, PointSymbol=none, PointName={}]{A}{B}{C}{O}

end{pspicture}

end{document}

to get:

Here you basically define (atleast) three-points in space and let the circle pass through them.

Update 1:

To get the desired results, define some pseudo points (as in the so-called Bezier's trick and 3 tangential points on the circle as in

%&pdflatex

% !TeX TXS-program:compile = txs:///pdflatex/[--shell-escape]

documentclass[a4paper, pdf, x11names]{standalone}

usepackage{pstricks}

usepackage{pstricks-add, auto-pst-pdf}

%usepackage{pst-solides3d}

% https://tex.stackexchange.com/questions/7199/can-pstricks-or-others-draw-the-4-common-tangent-lines-of-2-disjoint-circles-w --> some help from Herbert's solution!

usepackage{pst-eucl}

begin{document}

    begin{pspicture}

        %set a few nodes at desired locations and employ the Bezier trick

    pnodes(-1.8,2){M1}(3,4.2){M2}(2,-1.5){M3}(-1.7, 3){M4}

    psdots(M1)(M2)(M3)

    pstCircleABC[CodeFig=true, CodeFigColor=white,linecolor=red, PointSymbol=none, PointName={}]{M1}{M2}{M3}{M4}{O}

    pstTriangle(0,4){C}(0,0){A}(3,0){B} 

    pstMiddleAB{A}{B}{I1} pstArcOAB{I1}{A}{B}

    pstMiddleAB{B}{C}{I2} pstArcOAB{I2}{B}{C}

    pstMiddleAB{C}{A}{I3} pstArcOAB{I3}{C}{A}

    end{pspicture}



end{document}

to get:

PS: Of course this solution is not sophisticated, but this can be achieved without too much calculations or whatsoever. Only, some trial-and-error to place the points is required.