Find a vector parametrization of the circle contained in the plane x=5 with radius 3 centered at the point (5,1,2)

Question

I know the vector parametrization of the circle contained within the yz-plane centered at the origin is:

$\vec r$( $\theta$ ) = < 0, 3 cos $\theta$ , 3 sin $\theta$ >

What do I do next? If the circle was contained in the xy-plane or xz-plane, how would this change the parametrization?

It's worth noting that this question as stated doesn't quite make sense; a circle contained in the plane $x=5$ has its center in the same plane, so its center could never be $(3,1,2)$. @AbdallahHammam's answer is one reasonable interpretation, but note that the resulting circle doesn't have radius $3$. — stewbasic, Oct 17 '16 at 00:48

score 2 · Accepted Answer · edited Nov 23 '18 at 04:41

2

The circle we want the parametrization for, which is the intersection of the sphere of equation

$$(x-5)^2+(y-1)^2+(z-2)^2=9$$

with the plane $x=5$ has equation

$$(y-1)^2+(z-2)^2=9.$$

Thus, we can parametrize the circle in the following way:

\begin{align} x&=5\\ y&=1+3\cos(t)\\ z&=2+3\sin(t).\\ \end{align}

edited Nov 23 '18 at 04:41

Alex D

answered Oct 16 '16 at 23:24

hamam_Abdallah

1 Answers1