How to find points of the edge of one sphere inside another sphere. I am highly unqualified for this kind of discussion, but I need it for a CS project that I am working on at the moment.
I have one large sphere and inside it is a smaller sphere. When this smaller sphere collides with the edge of the larger sphere, I want to get a grid of points that form the edge of the larger sphere in a relatively simplistic calculation (aka I don't want to test a large chunk of points to see if they form the edge). Here is a 2D version of what I want to do that I made in MS Paint:

 A: Following up on your answer to my comment, I assume you need your points to be "evenly" spaced out on the spherical cap, and the tricky part here is what precise meaning you put behind the word "evenly".
In essence, you want to define some sort of spherical grid on the large sphere, then clip it with the smaller sphere. The problem with defining such a grid is that the easy and quick way to do that would be to use longitude/latitude coordinates on the sphere, and evenly divide the angle parameters so as to obtain a grid. The main problem with that approach, is that around the north and south poles, the density of the grid is much higher than around the equator as you can see in this image from wikipedia:

(yes I was too lazy to make my own picture)
A better way to define a grid on a sphere would be to use a geodesic grid. Once again a wiki link, if you scroll down the page they list some pros/cons of geodesic grids compared to the above longitude-latitude grid.
Anyway, back to the "simple" case.
Let's call the larger sphere $s_0$, it is centered at $c_0$ and has radius $r_0$. Likewise for the smaller sphere we use the notations $s_1$, $c_1$ and $r_1$.
Using maths convention for spherical coordinates, $P$ belongs to $s_0$ if its coordinates can be written 
\begin{align*}
x_P&=x_{c_0}+r_0\sin\varphi\cos\theta \\
y_P&=y_{c_0}+r_0\sin\varphi\sin\theta \\
z_P&=z_{c_0}+r_0\cos\varphi
\end{align*}
When using a rectangular grid, you should center your grid around the equator to reduce anisotropy. The intersection between the two spheres define a circle, its center lies on the line joining the centers of the sphere, so we want the line $(c_0c_1)$ to intersect the equator of the grid. Using standard coordinates conversion, you can compute the parameters $\theta_1$ and $\varphi_1$ that correspond to the direction of line $(c_0c_1)$.
You can then rotate the whole thing to make $(c_0c_1)$ parallel to plane
$xOy$. Then take the points
$$\left(r_0,\ i\Delta\theta,\ \frac\pi2+j\Delta\varphi\right)$$
where $i,j\in\mathbb Z$, and $\Delta\theta$ and $\Delta\varphi$ are step values computed according to how many points you want, and rotate the whole thing back to its original position.
Just a heads-up, let $r$ be the radius of the circle of intersection between the two spheres. Assuming your small sphere is indeed always "small",
or more precisely that you only need values
$\lvert i\Delta\theta \rvert\le\frac\pi2$
and $\lvert j\Delta\varphi \rvert\le\frac\pi2$,
you need
\begin{align*}
\lvert \sin\left(j\Delta\varphi\right) \rvert &\le \frac r{r_0} \\
\cos\left(i\Delta\theta\right) &\le \frac r{\lvert\sin\left(j\Delta\varphi\right)\rvert r_0}
\end{align*}
for the point to be inside the small sphere.
One last thing, you can compute $r$ from $r_0$, $r_1$ and the distance between $c_0$ and $c_1$ using Pythagoras, so depending exactly what you want you should be able to tune the step values properly.
A: Let $r$ be radius of big circle and $s$ be radius of smaller. Assume big circle's middle is at origin and small circle's origin is at $(a,0)$ with $a>0$. In other words small circle's center is on positive half of $x$ axis.
Then the formulas for the circles are $r^2=x^2+y^2$ and $s^2=(x-a)^2+y^2$. To find the value of $x$ for where the circles intersect, then solve for $x$ using these two equations. Since $y^2=r^2-x^2$, substitute that into second equation to get $$s^2=(x-a)^2+r^2-x^2 \Rightarrow x = \frac{a^2+r^2-s^2}{2a} \tag{1}$$.
Then any value of $x$ between that computed using (1) and $r$ is in the range of the red arc (from OP). So points can be computed by plugging values of $x$ in that range into equation for big circle.
