The equations for $C_1, C_2 $ are
$ (x - x_1)^2 + (y - y_1)^2 = r_1^2 $
$ (x - x_2)^2 + (y - y_2)^2 = r_2^2 $
which has $6$ degrees of freedom (i.e. $6$ free parameters). However, by the conditions in the problem we must have
$ (x_A - x_1)^2 + (y_A - y_1)^2 = r_1^2 $
$ (x_B - x_2)^2 + (y_B - y_2)^2 = r_2^2 $
$ (x_1 - x_2)^2 + (y_1 - y_2)^2 = (r_1 + r_2)^2 $
The solution of the the first $2$ equations is
$C_1 = (x_1, y_1) = (x_A , y_A) + r_1 (\cos(\phi), \sin(\phi) ) $
$C_2 = (x_2, y_2) = (x_B, y_B) + r_2 ( \cos(\theta), \sin(\theta)) $
Angles $ \theta, \phi $ are selected such that the third equation is satisfied, i.e.
$ \| A - B + r_1 u_1 - r_2 u_2 \| = r_1 + r_2 $
where $u_1 = (\cos(\phi), \sin(\phi) ) , u_2 = (\cos(\theta), \sin(\theta) ) $
Squaring, and simplifying,
$ \| A - B \|^2 + 2 (A - B) \cdot (r_1 u_1 - r_2 u_2) - 2 r_1 r_2 (u_1 \cdot u_2) = 2 r_1 r_2 $
Noting that $u_1 \cdot u_2 = \cos(\phi - \theta) $, the above simplifies to
$ \| A - B \|^2 + 2 (A - B) \cdot (r_1 u_1 - r_2 u_2) + 2 r1 r2 (1 - cos(\phi - \theta) ) = 0 $$ \| A - B \|^2 + 2 (A - B) \cdot (r_1 u_1 - r_2 u_2) + 2 r_1 r_2 (1 - cos(\phi - \theta) ) = 0 $
So to find circles $C_1, C_2$:
Choose $r_1, r_2,$ and $\phi $
and solve the final equation for $ \theta $