Figure 6

Solution:

Let the radii of $c_1, c_2$ and $c$ be $r_1, r_2$ and $r$, respectively. Homothety of ratio $\frac{r}{r_1}$ with center $A_1$ takes circle $c_1$ to circle $c$ and points $E_1, F_1$ to points $B, D$, respectively. Thus it takes line $E_1F_1$ to line $BD$. Analogously, homothety of ratio $\frac{r}{r_2}$ with center $A_2$ takes line $E_2F_2$ to line $BD$. Consequently, lines $E_1F_1$ and $E_2F_2$ are parallel to line $BD$ (Fig. 6).

Furthermore, note that $|BE_1| \cdot |BA_1| = |BP| \cdot |BQ|$ and $|BE_2| \cdot |BA_2| = |BP| \cdot |BQ|$, implying $|BE_1| \cdot |BA_1| = |BE_2| \cdot |BA_2|$. Thus triangles $BE_1E_2$ and $BA_2A_1$ are similar and
$$\angle BE_1E_2 = \angle BA_2A_1 = \angle BDA_1 = \angle E_1F_1A_1 = \frac{1}{2} \angle E_1O_1A_1 = 90^\circ - \angle O_1E_1A_1, \quad (3)$$
whence
$$\angle E_2E_1O_1 = 180^\circ - \angle BE_1E_2 - O_1E_1A_1 = 90^\circ. \quad (4)$$
Analogously, $\angle E_1E_2O_2 = 90^\circ$. Hence the quadrilateral $E_1E_2O_2O_1$ is a right-angled trapezoid (or rectangle in the case $r_1 = r_2$) and the midpoint of the line segment $O_1O_2$ lies on the perpendicular bisector of the line segment $E_1E_2$, thus being equidistant from $E_1$ and $E_2$. Analogously, the midpoint of the line segment $O_1O_2$ is also equidistant from $F_1$ and $F_2$.

As line $O_1O_2$ is perpendicular to $BD$, line $O_1O_2$ is also perpendicular to $E_1F_1$. Thus the line segment $O_1O_2$ entirely lies on the perpendicular bisector of $E_1F_1$. This means that the midpoint of line segment $O_1O_2$ is equidistant from $E_1$ and $F_1$.

Altogether, we have shown that these four points lie on a circle with its center at the midpoint of the line segment $O_1O_2$.