Podemos suponer sin pérdida de generalidad que el círculo de $O$ es la unidad de círculo, y el círculo de $O'$ radio $r$ con su centro situado en la $(R,0)$. A continuación, por un punto $$A = (\cos \theta, \sin \theta)$$ 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 $R$, $r$ se muestra.
Otra animación que muestra el barrido de la legitimación como el punto de $A$ es elegido en $O$ se muestra.