<s>[INST] Find a plane whose intersection line with a hyperboloid is a circle Find a plane $\pi$ which involves x-axis and its intersection line with $$\frac{x^2}{4}+y^2-z^2=1$$ is a circle.

Because the plane want to be find involves x-axis,so set as $By+Cz=0$,then I must to determine its value such that $$\begin{cases}By+Cz=0\\\frac{x^2}{4}+y^2-z^2=1\end{cases}$$ is a circle in 3-dimensional space,apparently,$B\not=0$,so I replace $y$ with $-\frac{C}{B}z$ in $\frac{x^2}{4}+y^2-z^2=1$,then i get a elliptic cylinder,at last ,i neet to determine $B$ and $C$ such that the intersection line between plane and elliptic cylinder is a circle.$$\begin{cases}By+Cz=0\\\frac{x^2}{4}+\left[(\frac{C}{B})^2-1\right]z^2=1\end{cases}$$
but i can't go any further. 

thanks very much
 [/INST] Assume $\pi$ has an equation like $$by+cz=0$$ for some unknown $c$ and $b$. If $b=0$ then $\pi: z=0$ and then we have an intersection like $x^2/4+y^2=1$ which is an ellipse not an circle. The same story would be for $c=0$. Let $c\neq0,b\neq0$ and so $z=-by/c$ so the intersection would be $$(1/4)x^2+y^2-b^2y^2/c^2 = 1$$ or $$(1/4)x^2+\left(1-b^2/c^2\right)y^2 = 1$$ if we want to have a circle as an intersection we need: $$b^2/c^2=3/4$$ For example set $b=\sqrt{3},c=2$ and look at the following picture:

</s>