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I have:

eu = Exp[I  2 i \[Pi]/n];
m = 3; pts = {Re[eu], Im[eu]} /. n -> m;
circle = {Sin[u], Cos[u]};
center1 = Table[pts, {i, 0, m - 1}]; 
circle1 = Map[(# + circle) &, center1];
ParametricPlot[circle1, {u, 0, 2 \[Pi]}]

which produces this:

enter image description here

Now I want to find the coordinates of the intersection of these circles. I tried:

Solve[circle1[[1]] == (circle1[[2]] /. u -> u + 2 \[Pi] n) && 
n \[Element] Integers && u > 0 && u < 2 \[Pi], {u, n}]

but that did not return an answer.

  • Is Solve[] capable of solving trigonometric problems by specifying the range of the angle as I have done above or should I convert it to the Complex format?
  • When solving 1D equations or solving multiple equations with all the variables from the same domain, we can do Solve[expr,vars,dom]. How can we specify different domains for different variables? Is it ok to introduce them as additional equations as I have done above?

I am not too sure of the mathematics of my approach so I have also asked it on Maths stack exchange.

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2 Answers 2

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If you want the intersection of all three circles, write the x,y coordinates of all three circles about their centers in center1.

FindInstance[(x - center1[[1, 1]])^2 + (y - center1[[1, 2]])^2 == 1 
   && (x - center1[[2, 1]])^2 + (y - center1[[2, 2]])^2 == 1 
   && (x - center1[[3, 1]])^2 + (y - center1[[3, 2]])^2 == 1, {x, y}, 2]

{{x -> 0, y -> 0}}

To find the intersection between pairs of the circles, take them two at a time

FindInstance[(x - center1[[1, 1]])^2 + (y - center1[[1, 2]])^2 == 1 
  && (x - center1[[2, 1]])^2 + (y - center1[[2, 2]])^2 == 1, {x, y}, 2]
FindInstance[(x - center1[[1, 1]])^2 + (y - center1[[1, 2]])^2 == 1 
  && (x - center1[[3, 1]])^2 + (y - center1[[3, 2]])^2 == 1, {x, y}, 2]
FindInstance[(x - center1[[2, 1]])^2 + (y - center1[[2, 2]])^2 == 1 
  && (x - center1[[3, 1]])^2 + (y - center1[[3, 2]])^2 == 1, {x, y}, 2]

{{x -> 0, y -> 0}, {x -> 1/2, y -> Sqrt[3]/2}}
{{x -> 0, y -> 0}, {x -> 1/2, y -> -(Sqrt[3]/2)}}
{{x -> -1, y -> 0}, {x -> 0, y -> 0}}
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  • $\begingroup$ Thank you, but you are doing it in Cartesian coordinates. I want to be able to do it Polar coordinates, or at least know why I can't. Regardless, would you care to explain the difference in usage of Solve and FindInstance? $\endgroup$
    – Shb
    Commented Dec 21, 2015 at 1:51
  • $\begingroup$ I guess it seems simpler in Cartesian coordinates. It looks like you get pretty much the same thing with Solve as with FindInstance in this case. $\endgroup$
    – bill s
    Commented Dec 21, 2015 at 3:03
  • $\begingroup$ Well this is a general question. Btw, I just realized I can try something like: Solve[Norm[circle1[[j]]] == Norm[circle1[[i]]] && u > 0 && u <= [Pi], u] with i and j pairs of {1,2,3}. It gives the right answer for {1,2} and {3,1} but for {2,3} which should return pi, it gives me pi/2. Any idea why? $\endgroup$
    – Shb
    Commented Dec 21, 2015 at 3:39
  • $\begingroup$ Neat solution! Thank you for sharing. $\endgroup$
    – ktm
    Commented May 23, 2017 at 22:36
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eq = {a, b, c, d} = {2 {Cos[u], Sin[u]}}~Join~circle1;
fun[w_, v_] := (w /. u -> p) /. {ToRules@
    N[Reduce[(w /. u -> p) == (v /. u -> q) && 0 <= p <= 2 Pi && 
       0 <= q <= 2 Pi, {p, q}]]}
pts = Chop[Join @@ fun @@@ Subsets[eq, {2}]]
ParametricPlot[
 Evaluate[{2 {Cos[u], Sin[u]}}~Join~circle1], {u, 0, 2 \[Pi]}, 
 Epilog -> {Red, PointSize[0.02], Point[pts]}]

enter image description here

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