Track chord movement

Question

The point $P$ on the circle and the point $F$ inside the circle, perpendicular bisector of line $PF$ intersects the circle at two points $A, B$, the envelope of $AB$ is an ellipse. I use the following code for visualization, I wan't to know how to track AB in a simple and efficient way.

Manipulate[Module[{O, P, A, B, F},
  O = {0, 0};
  P = AngleVector[t];
  F = {0.4, 0.5};
  {A, B} = RegionIntersection[Circle[], PerpendicularBisector[{P, F}]][[1]];
  Graphics[{
    Circle[],
    {Line[{P, F}], Red, Line@{A, B}},
    {Point[{O, P, F}]}
    }, PlotRange -> 1.2]
    ], {t, 0., 2 Pi}, TrackedSymbols :> {t}]

Answer 1

Edit-2

Manipulate[
 Module[{F, o, P, AB},
  F = Rationalize[{0.4, 0.5}, 0];
  o = {0, 0};
  P[t_] = AngleVector@t;
  AB[t_] := 
   AngleVector@s /. 
    NSolve[{(AngleVector@s - F) . (AngleVector@s - 
          F) == (AngleVector@s - P@t) . (AngleVector@s - P@t), 
      0 <= s <= 2 π}, s];
  Show[Table[
    Graphics[{{Hue[Rescale[t, {0, 2 π}]], 
       Line[AB@t]}}], {t, .1, τ, .1}], 
   Graphics[{Circle[], Point[{o, F}], Line[{F, P@τ}], 
     Line[AB@τ]}], PlotRange -> 1.2]], {τ, 0, 2 π}]

Edit-1

Use the same setting of the questionor.

Clear[line, plot, n, lines];
line[t_] := Module[{O, P, A, B, F}, O = {0, 0};
   P = AngleVector[t];
   F = {0.4, 0.5};
   {A, B} = 
    RegionIntersection[Circle[], PerpendicularBisector[{P, F}]][[1]];
   Graphics[{Hue[Rescale[t, {0, 2 π}]], 
     Line@{A, B}}, PlotRange -> 1.2]];
plot[t_] := Module[{O, P, A, B, F}, O = {0, 0};
  P = AngleVector[t];
  F = {0.4, 0.5};
  {A, B} = 
   RegionIntersection[Circle[], PerpendicularBisector[{P, F}]][[1]];
  Graphics[{Circle[], {Line[{P, F}], 
     Line@{A, B}}, {Point[{O, P, F}]}}, PlotRange -> 1.2]]
n = 40;
lines = Table[line[t], {t, Subdivide[2 π, n]}];
Manipulate[Show[lines[[1 ;; j]], plot[j/(2 π)]], {j, 0, n, 1}]

Edit-0

Need to be improve since the RegionIntersection too slow in my cases.

F = Rationalize[{0.4, -0.5}, 0];
o = {0, 0};
P[t_] = AngleVector[t];
e[t_] = λ*P@t /. 
   Simplify[
     Solve[{EuclideanDistance[λ*P@t, F] == 1 - λ, 
       0 <= λ <= 1, 0 <= t <= 2 π}, λ, Reals], 
     0 <= t <= 2 π][[1, 1]];
c[t_] = Midpoint[{P@t, F}];
plot[t_] = 
  Graphics[{Circle[], Point[{o, F}], Point[{e@t, c@t}], Red, 
    Point[P@t], {Green, 
     Line[{{e@t, o}, {e@t, P@t}, {e@t, F}}]}, {Dotted, 
     Line[{F, P[t]}]}}];
n = 40;
lines = Table[
   Graphics[{{ColorData["Rainbow"][Rescale[t, {0, 2 π}]], 
      RegionIntersection[Disk[], InfiniteLine[{c[t], e[t]}]]}}], {t, 
    Subdivide[2 π, n]}];

Manipulate[
 Show[plot[(j - 1)/(2 π)], lines[[1 ;; j]]], {j, 1, n, 1}]

Answer 2

bag = <||>; With[{O = {0, 0}, P = AngleVector[t], F = {0.4, 0.5}}, 
 Manipulate[DynamicModule[{A, B},
   If[! KeyExistsQ[bag, t],
    {A, B} = RegionIntersection[Circle[], PerpendicularBisector[{P, F}]][[1]];
    AppendTo[bag, t -> {Hue[t/(2 Pi)], Line@{A, B}}]];
   Graphics[{Circle[], {Line[{P, F}], Values@bag}, {Point[{O, P, F}]}}, 
    PlotRange -> 1.2]], {t, 0, 2 Pi, 2 Pi/100}, TrackedSymbols :> {t}]]

Answer 3

We assume a unit circle at the origin. Given a off center point: p0 and a point on the circle: {Cos[phi],Sin[phi]}, pm is the mid-point between these two points. Now A and B are determined by the fact that pm-{x,y} and pm-p0 are orthogonal. From this we can calculate all A and B:

p0 = {0.4, .4}; (*given off center point*)
pc[phi_] = {Cos[phi], Sin[phi]}; (*point on circle*)
pm[phi_] = p0 + (pc[phi] - p0)/2; (*midpoint between pc and p0*)
a[phi_] := 
  Solve[{{x, y} . {x, y} == 1, (pm[phi] - {x, y}) . (pm[phi] - p0) == 
      0}, {x, y}] // Quiet;
Graphics[{Circle[], 
  Line[Table[{x, y} /. a[phi], {phi, 0, 2 Pi, Pi/20}]], Red, 
  Point[p0]}]