Points on a Parameterized Circle in R3 I'm trying to solve a geometry problem that involves the intersection of two arbitrary spheres in 3D space. The intersection between the two spheres will be a circle in 3d space. I know that this circle will be contained in a plane with normal vector $\vec{a}$, defined as the vector pointing from the center of one sphere to the center of the other, and a point $M$, which is the centre of the circle of intersection, and will be some distance $d$ along $\vec{a}$ (this isn't the meat of the problem, I'm just giving some background). Finally, I find an orthonormal basis for my plane made up of vectors $\hat{u}$ and $\hat{v}%$ ($\hat{u}$ and $\hat{v}%$ have magnitude 1, and are both orthogonal to each other, and $\vec{a}$).
The circle of intersection can then be defined as the following:
$$C(x,y,z) = M + h*cos(t)*\hat{u} + h*sin(t)*\hat{v} $$
where $0 \leq t \leq 2\pi $
So now I have a parameterization of my circle, parameterized to some angle t, and this is where my question really starts. For my problem, I need to find certain points on $C(x,y,z)$ that obey a constraint in either x or y. For example, I may want to find all points in $C(x,y,z)$ such that $x = 5$, or such that $x$ is equal to some arbitrary number $b$. To do this, I look at only one variable in the above equation:
$$C(x) = b = M_x + h*cos(t)*\hat{u_x} + h*sin(t)*\hat{v_x}$$
where $M_X, \hat{u_x}, \hat{v_x}$ are the x components only. I rearrange the equation to look like:
$$\frac{b - M_x}{h} =cos(t)*\hat{u_x} + *sin(t)*\hat{v_x}$$
and then, for simplicity, let $k = \frac{b - M_x}{h}$:
$$k =cos(t)*\hat{u_x} + *sin(t)*\hat{v_x}$$
To solve this equation, I've been using the method described in this post, and it's been yielding correct solutions. The problem is that if I use the method described in that post, it will only give me one solution, when usually there will be two possible solutions for $t$ that satisfy my equation. What modifications can I make to my method such that I find all possible values for t, instead of just one? I have a matlab script running that does this calculation for me, which I can post if that would be helpful. I'd prefer not to use numerical methods in my code, as I'm sure this has an analytical solution.
 A: The method will give you both solutions. You've just been ignoring the other one. Let's look at it.
First we calculate $r = \sqrt{(\hat u_x)^2 + (\hat v_x)^2}$. Then we find $\phi$ so that $\cos \phi = \frac{\hat u_x}r$ and $\sin \phi = \frac{\hat v_x}r$
Then we can divide through by $r$ and rewrite the equation as $$\frac{b-M_x}{rh} = k = \cos t \cos \phi + \sin t \sin \phi = \cos(t - \phi)$$
where I have replaced your $k$ with one divided by $r$. Let $\theta = t - \phi$. Then the equation is $$k = \cos\theta$$
But recall the shape of the cosine curve. In $[0,2\pi)$ it drops from $1$ at $\theta = 0$ down to $-1$ at $\theta = \pi$, then rises back up to $1$ at $\theta = 2\pi$. With the exception of $-1$ at the bottom of the trough and $1$ at the top, the curve passes through every $y$-value twice! ($2\pi$ doesn't count as a second $1$ because it is actually in the next period.) So unless your $k$ is $1$ or $-1$, there are two values in $[0,2\pi)$ that provide it. By the symmetry of the cosine, $$\cos \theta = \cos (2\pi - \theta)$$
So once you find one solution, subtract it from $2\pi$ to get the other:
Suppose $k = \frac 12$, then $\theta = \arccos \frac12 = \frac \pi 3$. The other solution is then $2\pi - \frac \pi 3 = \frac{5\pi}3$.
Then just add $\phi$ to both solutions to get the values of $t$ you need.
