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In 3d space i have a circle in the $xz$ plane defined by equation $(x-i)^2+(z-k)^2=r^2$. If this circle is rotated through $\theta$ degrees around the line $z=k$ or by $\phi$ degrees around the line $x=i$ or both. What now is the equation?

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For a rotation around $z=k$, the resulting circle is the intersection between the sphere $(x-i)^2+(z-k)^2+y^2=r^2$ and the plane $y=\tan\theta(z-k)$.

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  • $\begingroup$ does it also hold for $x=i$ is the intersection of sphere and plane $y=tan\phi(i-x)$ $\endgroup$
    – rhubarbdog
    Jul 20 '16 at 4:50
  • $\begingroup$ Yes it does. Be careful to adjust the sign depending on how you choose the rotation angle. $\endgroup$ Jul 20 '16 at 7:09
  • $\begingroup$ is there any simple way to combine both $\theta$ and $\phi$ into one equation for a plane $\endgroup$
    – rhubarbdog
    Jul 20 '16 at 19:03
  • $\begingroup$ What do you mean exactly? What is the result you need? $\endgroup$ Jul 20 '16 at 19:19
  • $\begingroup$ something like $y=tan \theta(z-k)+tan \phi(x-i)$ $\endgroup$
    – rhubarbdog
    Jul 20 '16 at 19:39

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