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Let $C$ denote the curve of intersection of the two given surfaces:

$x+y=2$ and $x^2+y^2+z^2=2(x+y)$.

find a parametrization to $C$.

Here's my solution attempt:

Take a parameter $t\in\mathbb{R}$ and set $y=t$. Hence it follows that:

$x=2-t$ and that

$4-4t +2t^2+z^2=4$ $\implies$ $z^2=4t-2t^2$.

Therefore our parametrization is $\gamma(t)=(2-t, t ,\sqrt{4t-2t^2})$.

Is it correct? what happens with the negative values of z?

user2345678
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    Your parametrization does not "hit" the entire curve of intersection. (Separately, your formula only makes sense for $0 \leq t \leq 2$.) Just asking: Do you know (geometrically) what the intersection curve is? – Andrew D. Hwang Oct 30 '16 at 16:34
  • No, i do not kow what this curve actually is. Why did you argue that this parametrization doesn't hit the entire curve? thanks. – user2345678 Oct 30 '16 at 16:37
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    Thiking a little bit, i guess that the intersection is a circunferece, since $x^2+y^2+z^2=2(x+y)$ is a sphere and $x+y=2$ is a plane. – user2345678 Oct 30 '16 at 16:40
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    Precisely. And your curve traces a semicircle. :) – Andrew D. Hwang Oct 30 '16 at 16:48
  • So, the final response would be separate the curve of intersection into two parts:

    $\gamma_1(t)$ for z positive, and $\gamma_2(t)$ for z negative. The range of t would still be the same $t\in[0,2]$.

    Right?

     – user2345678 Oct 30 '16 at 16:51
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    Parametrizing by two semicircles is one possibility. Another would be to use trig functions, see for example the accepted answer to Parametrization for intersection of sphere and plane (the sphere differs from yours, but by chance the intersection curve is identical). – Andrew D. Hwang Oct 30 '16 at 17:00

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