I need a way to find the center of a circle of a fixed size nestled tangentially on the inside intersection of two other circles of fixed size and distance, as well as its points of intersection.
In the image, I have the radii of all 3 circles, as well as the distance between the larger circles A and B. The problem is finding the center of the smaller circle C and the intersecting points Q and P.
Note that $C$ belongs to the segment $\overline{AQ}$ since both $\overline{AQ}$ and $\overline{CQ}$ must be perpendicular to the tangent line at $Q$. By a similar argument, $C$ belongs to the segment $\overline{BP}$.
Therefore, $$ \overline{AC} = \overline{AQ} - \overline{CQ} = r_A - r_C \qquad\mbox{and}\qquad \overline{BC} = \overline{BP} - \overline{CP} = r_B - r_C $$ and $C$ is the intersection of two circles, one centred at $A$ with radius $\overline{AC}$, and another, centred at $B$ with radius $\overline{BC}$:
(Note: I'm using $\overline{X}$ to represent both the segment $X$ and its length)
$$ (x_C-x_A)^2 + (y_C-y_A)^2 = (r_A - r_C)^2 $$ $$ (x_C-x_B)^2 + (y_C-y_B)^2 = (r_B - r_C)^2 $$
These two equations can be solved for $(x_C,y_C)$. There will be, in general, two solutions. Choose either, as the other is mirrored with respect to $\overline{AB}$.
Once you have $(x_C,y_C)$, construct the unit vectors $$ \hat{u} = \frac{C-A}{\overline{AC}} $$ $$ \hat{v} = \frac{C-B}{\overline{BC}} $$
Then, $$ Q = r_A\,\hat{u} $$ $$ P = r_B\,\hat{v} $$
HINT:
A well-known circle theorem tells us that the tangent line to a circle at a point is at right angles to the radius ending at that point. If a small circle is tangent to two larger circles then you only need to find the radii of the larger circles at the tangent points and see where their cross.
(You might need to extend the radii into lines to find the intersection.)
