I know the dimensions of all the circles, but I'm unsure how to exactly get the points on a bigger circle - where all the smaller circles will neatly stack up.
Imgur Sketch: big circle w/ smaller circles.jpg
Hopefully the explanation was clear enough.
On
Put $n$ equally spaced points on the circumference of a circle of radius $r$ at points with coordinates $$ (r \cos(2j\pi/n), r \sin(2j \pi/n) ) $$ for $j = 0, \ldots, n-1$.
Figuring out the radius of the small circles is a little tricker - you say you have that number. Or see @amd 's answer for how to compute it. There $\theta = 2\pi/n$.
Two tangent circles of the same radius intersect at the midpoint of the line segment that joins their centers. If we call the radius of the small circles $r$ and the radius of the big circle $R$, this means that the length of the chord joining adjacent small circle centers is equal to $2r$, and so the angle $\theta$ between them satisfies $$\sin\frac\theta2 = \frac r R.$$ With this in hand, computing the coordinates of thse points is straightforward.
Geometrically, pick a point on the big circle as the center of your first little circle and draw a circle of radius $2r$ centered on this point. The intersection of this circle with the big one gives you the centers of the two adjacent small circles. Repeat in either or both directions until you’ve run out of room.
Analytically, you could reproduce the above process, but it’s much simpler to use rotations to generate the other centers from a given starting point.