# Create region from set of lines and arcs [duplicate]

I have a shape that is a knuckle plate with two holes cut in it, thus:

center1 = {0, 0};
center2 = {8, 0};
arc1 = Circle[center1, radius1, { -1 angle1, 2 \[Pi] + angle1}];
arc2 = Circle[center2, radius2, { -1 angle2, angle2}];
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];
angle1 = ArcTan[tangent[[1, 1, 2]]/tangent[[1, 1, 1]]];
angle2 = ArcTan[tangent[[2, 1, 2]]/tangent[[2, 1, 1]]];
bores = Circle[#, 1] & /@ {center1, center2};
Graphics[{arc1, arc2, Line[tangent], bores}]


I want to be able create a region which is defined by the perimeter of the knuckle plate subtracted by the holes. I started to do this and ran into problems:

plate = RegionUnion[arc1, arc2, Line[tangent]];
platemesh = BoundaryDiscretizeRegion[plate];
Graphics[{platemesh}]


This gives errors such as "There is not a boundary representation that uniquely defines a region with region dimension 1 embedded in dimension 2". (Note that I am hoping for a general answer that would work for any set of connected lines and arcs, not just a convex shape.)

Note on potential duplicates:

Solutions that involve ConvexHull are not applicable as explained above

Solution from a related post:

Polygon@points[[First@Rest@FindShortestTour@points]]


Involves flattening of the line segments into discrete points, which for a complex set of paths could be complicated. Ideally, I am looking for a solution that computes the region using the paths as the primary inputs without having to break those paths into more primitive objects.

## marked as duplicate by MarcoB, Dr. belisarius, dr.blochwave, Mr.Wizard♦Aug 14 '15 at 15:24

• It would seem to me that the answers you received and contributed yourself to your very similar previous question from yesterday should apply here as well. – MarcoB Aug 13 '15 at 15:15

I am using all your definitions, except for bores, which I modified to use Disk instead of Circle:

bores = Disk[#, 1] & /@ {center1, center2};

RegionDifference[
DiscretizeRegion@RegionUnion[