Find min and max values of circle coordinates in space Is there a way to find $\min$ and $\max$ values of $x$, $y$, and $z$ coordinates of a circle in $\mathbb{R}^3$? The radius of, center of, and plane containing the circle are given (the plane is given by a normal vector and a point).
 A: As noted in the comments, we will solve this question for a unit circle centered at the origin.

Using the technique presented in this answer, we will find a set of parametric equations describing the circle.
If $\mathbf{v}$ is the unit normal vector to the plane containing the center, we find a vector in the plane by solving $\mathbf{a} \cdot \mathbf{v} = 0$ for $\mathbf{a}$. Then we can take the cross product to get a vector in the plane that is orthogonal to $\mathbf{a}$:
$$\mathbf{b} = \mathbf{a} \times \mathbf{v}$$
Make sure at this point that $\mathbf{a}$ and $\mathbf{b}$ are properly normalized so that they both have length 1 (otherwise we would get an ellipse).
Then we can represent the circle using the following parametric equations:
$$x(\theta) = \cos(\theta)a_1 + \sin(\theta)b_1 \,,$$
$$y(\theta) = \cos(\theta)a_2 + \sin(\theta)b_2 \,,$$
$$z(\theta) = \cos(\theta)a_3 + \sin(\theta)b_3 \,,$$
where $\mathbf{a} = (a_1,a_2,a_3)$ and $\mathbf{b} = (b_1,b_2,b_3)$.
Since the functions are differentiable, we know that their maxima and minima will be attained at points where their derivatives are zero. We take the derivatives:
$$x'(\theta) = -\sin(\theta)a_1 + \cos(\theta)b_1 \,,$$
$$y'(\theta) = -\sin(\theta)a_2 + \cos(\theta)b_2 \,,$$
$$z'(\theta) = -\sin(\theta)a_3 + \cos(\theta)b_3 \,.$$

The max and min for the $x$-coordinate will occur when $\tan(\theta) = \frac{b_1}{a_1}$.
The max and min for the $y$-coordinate will occur when $\tan(\theta) = \frac{b_2}{a_2}$.
The max and min for the $z$-coordinate will occur when $\tan(\theta) = \frac{b_3}{a_3}$.
