Parameterize a union of circles

[chipotleaway]

Homework Statement

Let 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. Give a parameterization of the curve C.

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 r(t)=(cos(t+\frac{\pi}{6}), sin(t+\frac{\pi}{6}) for all t \in [0, 2\pi) and r(t)=(cos(t+\frac{2\pi}{3})+1, sin(t+\frac{2\pi}{3}) for all t \in [2\pi, 4\pi].

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.

 

[SammyS]

Homework Statement

Let 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. Give a parameterization of the curve C.

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 r(t)=(cos(t+\frac{\pi}{6}), sin(t+\frac{\pi}{6}) for all t \in [0, 2\pi) and r(t)=(cos(t+\frac{2\pi}{3})+1, sin(t+\frac{2\pi}{3}) for all t \in [2\pi, 4\pi].

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.

What are the coordinates of the points of intersection ?

 

[haruspex]

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)?

 

[chipotleaway]

@SammyS: The circles intersect at x=\frac{1}{2}

@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?