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It is easiest to use ParametricPlot and RotationTransform. ParametricPlot[ RotationTransform[a][{1, 0}] + RotationTransform[4 a][{1/4, 0}], {a, 0, 2 Pi}, Evaluated -> True] Evaluated -> T …

Somewhat dumb method (for instance, every line has both two Disks and a StadiumShape overlapping), but it's not at least very complicated: hulls0 = ConvexHullMesh /@ RandomReal[{-10, 10}, {3, 1, 2}]; …

Based on your image, but not your statement on lengths of lines: With[{eqn = a - a x / (1 - a)}, Show[ Quiet@Plot[Table[eqn, {a, 0, 1, 1 / 20}], {x, 0, 1}, Evaluated -> True, AspectRatio -> Aut …

You can find a geometric solution to this problem by using an implicit region definition directly based on the definition of an ellipse: sum of distances from foci equals twice the semimajor axis. Wi …

Turning my comment into an answer per (now deleted?) comment which requested it. This is documented to work only in Wolfram Language at this point (specifically Wolfram Programming Cloud). Interestin …

This solution is primarily a proof of concept; its run-time complexity makes it impractical for large amounts of items. Large amount means more than six, in this case. Nonetheless, the idea is the fo …

There are numerous ways to do this in Mathematica, and it's hard to say which would be most useful for learning. Here's one; a unit circle is drawn, then a polygon with no filling and black edge on ba …

Correct me if I'm wrong... but I suppose, pedantically speaking, there are only five solutions defined by Platonic solids, and trivial solutions for 0-3 points (extension of CirclePoints). Thus: Cle …

If more ad hoc, inexact approaches are welcome, one way to generate relatively uniform density of points on a sphere is to use Monte Carlo Lloyd's algorithm (modified for the spherical case): With[{p …

EDIT: As @nikie noted, using FindArgMin (a variant of FindMinimum) instead of (N)ArgMin can improve the speed of finding a solution. Since in the case of this problem only one minimum exists, this sho …

Mathematica 11.1 has a built-in function for this task: SpherePoints. This roughly corresponds CirclePoints for three-dimensional case; thus I wouldn't rely on it to get any sort of randomness on the …

Assuming I understood your question correctly (there's really no need to use Module), here's one method which delegates calculating intersections to Solve: ClearAll@circleChord; circleChord[{xc_, yc …

I have the following problem, which can be stated geometrically as follows: List all pairs of vertex coordinates in a mesh region of measure 2 (a surface consisting of triangles) where these points a …

If you want to simply sample the intersection, you can do this: RandomPoint[ ImplicitRegion[ RegionMember[ RegionIntersection[Sphere[{0, 0, 0}, 1], Sphere[{1, 1, 1}, 1.5]], {x, y, …

This solution is a little bit... elaborate for the task, but shows how one can use GeometricTest to find solutions using synthetic geometry tools: With[ {c1 = {0, 0}, r1 = 1, c2 = {3, 4}, r2 = 4, …