Let $L_1$ and $L_2$ be the points of tangency of the external common tangent of the circles, $N_1$ and $N_2$ be the points of tangency of an internal common tangent, and $O_1$ and $O_2$ be the centers of the two circles (see Fig. 18).

As all the lines $O_1L_1$, $AK$, and $O_2L_2$ are perpendicular to the line $L_1L_2$, they are parallel to each other and thus $$\frac{|L_1K|}{|L_2K|} = \frac{|O_1A|}{|O_2A|}.$$ The triangles $O_1AN_1$ and $O_2AN_2$ are similar because they are both right-angled and have the same vertical angles. Thus, $$\frac{|O_1A|}{|O_2A|} = \frac{|O_1N_1|}{|O_2N_2|} = \frac{|O_1L_1|}{|O_2L_2|}.$$ Therefore, the right-angled triangles $O_1L_1K$ and $O_2L_2K$ are similar due to proportionality of their legs. Hence, $\angle L_1KO_1 = \angle L_2KO_2$.

As $\angle L_1KM_1 = 2\angle L_1KO_1$ and $\angle L_2KM_2 = 2\angle L_2KO_2$, we also get that $\angle L_1KM_1 = \angle L_2KM_2$. Together with the equality $\angle L_1KA = \angle L_2KA = 90^\circ$ this implies $\angle M_1KA = \angle M_2KA$.

Both of the circles appear at the same angle, when viewed from the point $A$. To solve the problem, it is enough to show that both of the circles also appear at the same angle, when viewed from the point $K$.

Let the centers of the circles have the coordinates $O_1(a_1, b_1)$ and $O_2(a_2, b_2)$ and let $r_1$ and $r_2$ be the radii of the circles. The two circles appear at the same angle from the point $P(x, y)$ if and only if $$\frac{r_1}{|O_1P|} = \frac{r_2}{|O_2P|},$$ i.e., $$\frac{r_1}{\sqrt{(x-a_1)^2+(y-b_1)^2}} = \frac{r_2}{\sqrt{(x-a_2)^2+(y-b_2)^2}}.$$ Simple algebra shows that this equation is equivalent to $$(r_1^2 - r_2^2)x^2 + (r_1^2 - r_2^2)y^2 + c_1x + c_2y + c_3 = 0,$$ where $c_1, c_2$, and $c_3$ are some constants.

If $r_1 = r_2$, then the statement clearly holds. If $r_1 \neq r_2$, then the last equation is that of a circle. Point $A$ as well as the point $D$ of intersection of the external common tangents both lie on that circle, and from symmetry, the diameter of that circle is $AD$. As $AK$ is perpendicular to the external common tangent of the circles, the point $K$ also lies on that circle.

Fig. 18