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Parameterize a union of circles

  1. Feb 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Let [itex]C=\lbrace(x,y) \in R^2: x^2+y^2=1 \rbrace \cup \lbrace (x,y) \in R^2: (x-1)^2+y^2=1 \rbrace [/itex]. Give a parameterization of the curve C.

    3. The attempt at a solution
    I'm not sure how valid it is but I tried to use a 'piecewise parameterisation', defining it to be [itex]r(t)=(cos(t+\frac{\pi}{6}), sin(t+\frac{\pi}{6})[/itex] for all [itex]t \in [0, 2\pi)[/itex] and [itex]r(t)=(cos(t+\frac{2\pi}{3})+1, sin(t+\frac{2\pi}{3})[/itex] for all [itex]t \in [2\pi, 4\pi][/itex].

    So it traces out the first circle out at the top intersection and then the second. But I'm not sure how to make this smooth at the intersection, and one-to-one, as I'm quite sure it has to be. This is the best I could come up with.
  2. jcsd
  3. Feb 2, 2014 #2


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    What are the coordinates of the points of intersection ?
  4. Feb 2, 2014 #3


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    You would parameterise the one circle as x = r cos(t), etc., and the other as x = 1+r cos(t). How would you write the condition that x is either r cos(t) or 1+r cos(t)?
  5. Feb 13, 2014 #4
    @SammyS: The circles intersect at [itex]x=\frac{1}{2}[/itex]

    @haruspex: As a hybrid function/parameterization? So x=cos(t) for t in some interval and x=cos(t)+1 for t in another interval?

    Would this mean we are doing two separate parameterizations and so they don't have to be smooth at the intersection?
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