Well, you might use an algebraic approach; I'll help you along a bit:
Let the radii be r and R, respectively.
Let the r-circle have centre at (0,0), the R-circle at (R+r,0)
Thus, the two circles have the representations:
x^{2}+y^{2}=r^{2}, (x-(r+R))^{2}+y^{2}=R^{2}
Let a point of tangency at the r-circle be denoted as (x_{0},y_{0})
(Note that the x-value of the tangency points on the r-circle will be unique, whereas there must be two y-values).
The unit tangent at (x_{0},y_{0}) in the direction of the other circle is \vec{T}=(\frac{y_{0}}{r},-\frac{x_{0}}{r}), and we may parametrize its tangent line as:
\vec{L}(t)=(x_{0},y_{0})+t\vec{T}
Now, put this expression into the equation for the R-circle:
(x_{0}+\frac{y_{0}t}{r}-(r+R))^{2}+(y_{0}-\frac{x_{0}t}{r})^{2}=R^{2}
Or, we may rewrite to:
t^{2}-\frac{2y_{0}(r+R)}{r}t+2(r-x_{0})(r+R)=0
which is a second degree equation in t, as expressed with the other variables!
Now, note that if \vec{L} is also tangent to some point on the R-circle, such a point will be the ONLY point on the R-circle coinciding with \vec{L}
If \vec{L} is NOT a tangent line to the R-circle, it will either intersect this circle twice, or not at all.
Thus, if we require that the discriminant of the second-degree equation is zero, then we know that the t-value thus gained gets us the point of tangency on the R-circle.
The requirement that the discriminant be zero, furnish us with the last equation in order to determine (x_{0},y_{0}) in terms of r and R (the other being the trivial requirement that the point lies on the r-circle.
Hope this helps.
EDIT:
In the above, I have of course ignored that the two circles have a common tangent line also for their point of contact. You'll find this solution as well when going through the calculations.
