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I am reproducing the derivation in this article (exploration of a family of circles centered on an ellipse).
enter image description here enter image description here
I only need the outer curve, the results obtained from Mathematica are more complicated than Maple, and it's difficult to simplify

poly = (x - 4 Cos[t])^2 + (y - 3 Sin[t])^2 - 1;
eqn = {poly, D[poly, t]} == 0
{x, y} /. Solve[eqn, {x, y}] // Simplify[#, 0 < t < 2 Pi] &
ParametricPlot[%[[1]], {t, 0, 2 Pi}]

enter image description here

(*result from maple*)
ParametricPlot[{Cos[t] (3/Sqrt[16-7 Cos[t]^2]+4),Sin[t] (4/Sqrt[16-7 Cos[t]^2]+3)},{t,0,2Pi}]

enter image description here
Can I get the results similar to the Maple?

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  • $\begingroup$ eqn = {poly==0, D[poly, t]==0} ; Assuming[ 0 < t < Pi, FullSimplify[{x, y} /. Solve[eqn, {x, y}]] ] The solution for $\pi<t<2\pi$ is the same, both equal to whet you have in the document. $\endgroup$
    – rhermans
    Commented May 5, 2022 at 15:43
  • $\begingroup$ @rhermans It may depend on the version, I'm using version 12.0, can't get the results in the documentation. $\endgroup$
    – expression
    Commented May 5, 2022 at 16:00
  • $\begingroup$ I get two branches, with an unfortunate branch cut on the y-axis, in V13: i.sstatic.net/DiBt8.png $\endgroup$
    – Michael E2
    Commented May 5, 2022 at 17:11

1 Answer 1

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I had to play with the domain of t to fix the branch-cut switch (note that -Pi < t <= Pi is the range of t == ArcTan[_, _], which may or may not be relevant but was why I tried it):

poly = (x - 4 Cos[t])^2 + (y - 3 Sin[t])^2 - 1;
eqn = {poly == 0, D[poly, t] == 0, -Pi < t < Pi};
sols = {x, y} /. Solve[eqn, {x, y}] // Simplify[#, 0 < t < 2 Pi] &;

ParametricPlot[sols, {t, -Pi, Pi}]

enter image description here

Thus sols[[2]] is the outer loop.

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