= Solution
The <plane equation> $x+z=0$ has <normal vector> $(1,0,1)^T=\sqrt2\hat{\mathbf n}$. Thus its <normal vector> is parallel to the <cylinder>'s axis and the <plane> is perpendicular to that axis. It passes through the origin, so the intersection is \b[a <circle> of radius $2$ centred at the origin in that <plane>].
Indeed, with the <orthonormal basis> $\mathbf e=(1,0,-1)^T/\sqrt2$ and $\mathbf f=(0,1,0)^T$ of the <plane>, write $\mathbf r=s\mathbf e+t\mathbf f$. The <cylinder> equation becomes $s^2+t^2=4$. Equivalently, the intersection has the <parametric equation>
$$
\boxed{\mathbf r(\theta)=2\cos\theta\,\mathbf e+2\sin\theta\,\mathbf f,\quad0\leq\theta<2\pi.}
$$
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