Given the circle, midpoint $O(0,0)$ and radius $K\ :\ x^{2}+y^{2}=K^{2}$. Given the circle, midpoint $O_{1}(K-L,0)$ and radius $L\ :\ [x-(K-L)]^{2}+y^{2}=L^{2}$, point of tangency $(K,0)$. Construct the line $y=\tan \alpha \cdot x$. On this line, midpoint $O_{2}(L\cos \alpha,L\sin \alpha)$. Circle with midpoint $O_{2}$ and radius $K-L\ :\ (x-L\cos \alpha)^{2}+(y-L\sin \alpha)^{2}=(K-L)^{2}$, point of tangency $(K\cos \alpha,K\sin \alpha)$. The last two circles cut in $(L\cos \alpha+K-L,L\sin \alpha)$. This point lies on the line through the points of tangency $y=\frac{\sin \alpha}{\cos \alpha-1}(x-K)$.

solved synthetically here