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I have two circles and tangential line segments that I want to define a region:

center1={0,0};
center2={8,0};
radius1 = 4;
radius2 = 3;
circle1=Circle[center1,radius1];
circle2=Circle[center2,radius2];
endpoint1={x1,y1};
endpoint2={x2,y2};
tangent={endpoint1,endpoint2}/.Solve[{
(endpoint2-center2).(endpoint2-endpoint1)==0,
(endpoint1-center1).(endpoint2-endpoint1)==0,
(endpoint1-center1).(endpoint1-center1)==radius1^2,
(endpoint2-center2).(endpoint2-center2)==radius2^2,
(endpoint1-center1).(endpoint2-center2)>0},             
    {x1,y1,x2,y2},Reals]
Graphics[{circle1,circle2, Line[tangent]}]

enter image description here

How can I delete the inner arcs of the circles and create a region that is bounded by the outer lines and arcs of the shape?

I have Mathematica 10.1 so I can use the latest functions in Geometry.

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3 Answers 3

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ConvexHullMesh[
 Flatten[Table[{4 (a = {Cos[t], Sin[t]}), {8, 0} + 3 a}, 
   {t, 0, 2 π, 0.1}], 1]]

enter image description here

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I would suggest using Disk instead of Circle as a start; the following generates a Region object that should generate what you are looking for:

center1 = {0, 0};
center2 = {8, 0};
radius1 = 4;
radius2 = 3;

circle1 = Disk[center1, radius1];
circle2 = Disk[center2, radius2];
endpoint1 = {x1, y1};
endpoint2 = {x2, y2};

tangent = {endpoint1, endpoint2} /. 
   Solve[{(endpoint2 - center2).(endpoint2 - endpoint1) == 
      0, (endpoint1 - center1).(endpoint2 - endpoint1) == 
      0, (endpoint1 - center1).(endpoint1 - center1) == 
      radius1^2, (endpoint2 - center2).(endpoint2 - center2) == 
      radius2^2, (endpoint1 - center1).(endpoint2 - center2) > 
      0}, {x1, y1, x2, y2}, Reals];

points = Flatten[tangent, 1];

RegionPlot[
 RegionUnion[
  circle1, circle2, 
  Polygon@points[[First@Rest@FindShortestTour@points]]
 ],
 AspectRatio -> Automatic
]

Mathematica graphics

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  • $\begingroup$ MarcoB: How would you "know" where the intersection points are? $\endgroup$ Aug 13, 2015 at 0:19
  • $\begingroup$ @David Quite. I I misread the question. Thanks for pointing that out; I removed that part. $\endgroup$
    – MarcoB
    Aug 13, 2015 at 0:32
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I have found a simple solution which is to compute the outer circular arcs and use the arcs instead of the circles:

angle1 = ArcTan[tangent[[1, 1, 2]]/tangent[[1, 1, 1]]];
angle2 = ArcTan[tangent[[2, 1, 2]]/tangent[[2, 1, 1]]];
arc1 = Circle[center1, radius1, { -1 angle1, 2 π + angle1}];
arc2 = Circle[center2, radius2, { -1 angle2, angle2}];
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