Let $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$ be two unit vectors perpendicular to the direction of the axis and each other, and let $(c_1,c_2,c_3)$ be any point on the axis. (If ${\bf v} = (v_1,v_2,v_3)$ is a unit vector in the direction of the axis, you can choose ${\bf a} = (a_1,a_2,a_3)$ by solving ${\bf a} \cdot {\bf v} = 0$, scaling ${\bf a}$ to make $\|{\bf a}\| = 1$, then letting ${\bf b} = {\bf a} \times {\bf v}$.)
Then for any $r$ and $\theta$, the point $(c_1,c_2,c_3) + r\cos(\theta)(a_1,a_2,a_3) + r\sin(\theta)(b_1,b_2,b_3)$ will be at distance $r$ from $(c_1,c_2,c_3)$, and as $\theta$ goes from $0$ to $2\pi$, the points of distance $r$ from $(c_1,c_2,c_3)$ on the plane containing $(c_1,c_2,c_3)$ perpendicular to the axis will be traced out.
So the parameterization of the circle of radius $r$ around the axis, centered at $(c_1,c_2,c_3)$, is given by
$$x(\theta) = c_1 + r\cos(\theta)a_1 + r\sin(\theta)b_1$$
$$y(\theta) = c_2 + r\cos(\theta)a_2 + r\sin(\theta)b_2$$
$$z(\theta) = c_3 + r\cos(\theta)a_3 + r\sin(\theta)b_3$$
Best Answer
We are given that $(i): x = 1 + \frac{1}{t}$ and $(ii): y = t^{2}$:
From $(i): t = \frac{1}{x-1}$. Then it follows from $(ii)$ that $$y(x) = t^{2} = \frac{1}{(x-1)^{2}}$$