Trigonometric solution

Suppose that circle i has radius $r_i$, that $d$ is the distance between centers, and that we form a plane section including the two centers and $p$, one of the points where the circles intersect. Let the perpendicular from $p$ to the line joining centers have length $c$ while dividing that line into segments of length $a$ and $b$. By the Pythagorean theorem,

$$\begin{eqnarray*} a^2+c^2 & = & r_1^2 \\ b^2+c^2 & = & r_2^2, \end{eqnarray*}$$

whence we conclude

$$\begin{eqnarray*} a^2-b^2 & = & r_1^2 - r_2^2. \end{eqnarray*}$$

At the same time,

$$\begin{eqnarray*} a+b & = & d. \end{eqnarray*}$$

Manipulating these last two equations gives forms for $a+b$ and $a-b$ which in turn yield

$$\begin{eqnarray*} a & = & \frac{r_1^2-r_2^2+d^2}{2d}; \end{eqnarray*}$$

the formula for $b$ results from exchangiong $r_1$ and $r_2$. Interestingly, the spheres need not even intersect, but then $c$ would be imaginary:

$$\begin{eqnarray*} c^2 & = & \frac{2r_1^2d^2+2r_2^2d^2+2r_1^2r_2^2-r_1^4-r_2^4-d^4}{4d^2}. \end{eqnarray*}$$