<s>[INST] Parametrization for intersection of sphere and plane Given is the sphere $x^2 + y^2 + z^2 = 4$ and the plane $x + y = 2$ in $\mathbb R^3 $.
How can I find a parametrization for the intersection of the two?
 [/INST] You are in luck because one of the equations is linear.
Since $x+y=2$ it follows that $x=2-y$. We can substitute this into the first equation:
$$x^2+y^2+z^2=4 \implies (2-y)^2+y^2+z^2=4 \implies 2y^2-4y+z^2=0\,.
$$
We can "complete the square" on the first part to give:
$$2[(y-1)^2-1]+z^2=0 \implies 2(y-1)^2+z^2=2 \, . $$
Considering only the $yz$-plane then we have an ellipse in the $yz$-plane parametrised by $(y(\theta),z(\theta)) = (1+\cos\theta,\sqrt{2}\sin\theta)$. We also know that $x=2-y$, hence:
$$\gamma(\theta) = (1-\cos\theta,1+\cos\theta,\sqrt{2}\sin\theta)$$
would give a regular parametrisation of the set in question.
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