# Parametrization of the osculating circle to a space curve?

Find a parametrization of the osculating circle to r(t)= at t=0

So I found the center of the osculating circle by calculating the radius of curvature and the normal vector. I've also found the equation of the osculating plane in which the circle will lie, but I'm stuck on how to parametrize the circle on this plane.

For time's sake, here are the unit tangent, unit normal, and binormal vectors, the center of the osculating circle, and the equation for the plane:

T(t)= (1/sqrt(2))<-sin(7t),cos(7t),1> N(t)=<-cos(7t),-sin(7t),0> B(0)=<0,1/sqrt(2),1/sqrt(2)>

center of osculating circle at t=0, = <-1,0,0> equation for osculating plane at t=0, = (1/sqrt(2))y+(1/sqrt(2))z=0

at this step, I tried to find the equation of a sphere with the same center and radius as the osculating circle, which gave me (x+1)^2 + y^2 + z^2 =4 and then find the intersection of this sphere with the osculating plane. Solving for y in terms of z in the osculating plane equations yields y=-z, and when plugged into the equation for the sphere, yields (x+1)^2 + 2y^2 = 4. I don't know if i'm on the right track here, but this is as far as i've been able to get.

If anyone could help me out, or show me an entirely different way of approaching the problem, it would be greatly appreciated.

Thanks!

$\newcommand{\Vec}{\mathbf{#1}}$Generally, if $\Vec{v}_{1}$ and $\Vec{v}_{2}$ are orthogonal unit vectors in $\Vec{R}^{n}$, then $$\Vec{x}(t) = \Vec{p}_{0} + r\bigl((\cos t)\Vec{v}_{1} + (\sin t)\Vec{v}_{2}\bigr)$$ parametrizes the circle of center $\Vec{p}_{0}$ and radius $r$ in the plane spanned by $\Vec{v}_{1}$ and $\Vec{v}_{2}$.