Tangents of two circles, a problem in Jules Verne book Paris in the 20th century by Jules Verne presents the following problem (translation mine):

Given two circles $O$ and $O'$: From point $A$ on $O$, two tangents are drawn for $O'$; a line is drawn between the points which they touch [on $O'$]; a tangent is drawn at point $A$ for the circle $O$; the question is what is the intersection point of this tangent and the line that was drawn between the two points of touch on $O'$.

How would I solve this problem? I would also like to know if this is a famous problem (I assume it was in 1863, why else Verne would have included it?).
Edit: Diagram plotted by Moti. In the diagram, point $C$ is the original problem's $A$.

 A: Here is a diagram that should help you:

A: Let me add a few features to the diagram:
The secant line $\overline{T_1T_2}$ can be determined without explicitly finding the two points of tangency. Note that it will be perpendicular to the segment $\overline{AO'}$. By similar triangles, $AT_1:AQ::AO':AT_1$, and the Pythagorean theorem gives $AO'^2=AT_1^2+O'T_1^2$. Putting these together, along with the fact that $O'T_1$ is just the radius $R'$ of the circle, we get $$
{AQ \over AO'}={AO'^2-R'^2 \over AO'^2}.
$$ This will let you find the point $Q$. The slope $m'$ of the line $\overline{AO'}$ is easily computed from $A$ and $O'$, and the slope of $\overline{T_1T_2}$ is $-1/m'$. The slope of the tangent to the circle $O$ at $A$ is also easily computed from $A$ and the circle’s center. That should be enough information to determine the two red lines and find their intersection $P$ (if it exists). 
A: We can assume without loss of generality that circle $O$ is the unit circle, and circle $O'$ has radius $r$ with its center located at $(R,0)$.  Then for a point $$A = (\cos \theta, \sin \theta)$$ on $O$, we are interested in the equation of the line $\ell(\theta)$ passing through the tangents from $A$ to $O'$.  But this line is precisely the perpendicular to the line segment $\overline{AO'}$ through a point $A'$, where $A'$ is the inversion of $A$ through circle $O'$.  In other words, $A'$ satisfies $$AO' \cdot A'O' = r^2.$$  Since the line through $A$ and $O'$ is given by $$y = \frac{\sin \theta}{\cos \theta - R}(x - R),$$ which we can also parametrize as $$(x(t), y(t)) = O'(1-t) + At = (R(1-t) + t \cos \theta, t \sin \theta), \quad 0 \le t \le 1,$$ and $$\frac{A'O'}{AO'} = \frac{r^2}{(R-\cos\theta)^2 + \sin^2\theta},$$ the intersection point $P$ of $AO'$ and $A'O'$ corresponds to the choice $t = A'O'/AO'$; i.e. $$P = \left(\frac{R(1-r^2 + R^2) + (r^2 - 2R^2)\cos \theta}{1-2R \cos \theta + R^2}, \frac{r^2 \sin \theta}{1 - 2R \cos \theta + R^2}\right).$$  Thus $$\ell(\theta) \equiv y - P_y = \frac{R - \cos \theta}{\sin\theta}(x - P_x).$$  The locus of the intersection of $\ell(\theta)$ with the tangent line $$\tau(\theta) \equiv y - \sin \theta = -\cot \theta (x - \cos\theta)$$ is (after a considerable amount of algebra) $$\mathcal L(\theta;R,r) = \left(\frac{1-r^2}{R} + R - \cos \theta, -\left(\frac{1-r^2}{R} + R\right) \cot \theta + 2 \csc \theta - \sin \theta \right).$$  An animation for various choice of $R$, $r$ is shown.

Another animation showing the sweeping of the locus as the point $A$ is chosen on $O$ is shown.

