For three circles that are not placed on top of each other, construct a point from which all three circles are seen at the same angle.
Let the radii of the circles be $r_{1}, r_{2}, r_{3}$, and their centers be $O_{1}, O_{2}$, and $O_{3}$. First, we determine the set of points from which two circles are seen at the same angle.

![](https://cdn.mathpix.com/cropped/2024_05_02_d7e0f62eafce6a75fbfeg-1.jpg?height=398&width=637&top_left_y=262&top_left_x=728)

Figure 1

Suppose that from point $P$, the circles centered at $O_{1}$ and $O_{2}$ are seen at the same angle. Let the points of tangency of the tangents drawn from $P$ to the circles be $E_{1}^{1}, E_{1}^{2}, E_{2}^{1}$, and $E_{2}^{2}$ (Figure 1). Then $E_{1}^{1} P E_{1}^{2} \varangle = E_{2}^{1} P E_{2}^{2} \varangle$, and since $O_{1} P$ and $O_{2} P$ are the angle bisectors of these angles, $O_{1} P E_{1}^{1} \varangle = O_{2} P E_{2}^{1} \varangle$. Thus, the right triangles $O_{1} E_{1}^{1} P$ and $O_{2} E_{2}^{1} P$ are similar. The ratio of the corresponding sides is:

$$
\frac{O_{1} P}{O_{2} P} = \frac{O_{1} E_{1}^{1}}{O_{2} E_{2}^{1}} = \frac{r_{1}}{r_{2}}
$$

It is easy to see that the two circles are seen at the same angle from $P$ if and only if (1) holds. This means that $P$ lies on the Apollonian circle corresponding to the points $O_{1}$ and $O_{2}$ and the ratio $r_{1}: r_{2}$ (in the case $r_{1} = r_{2}$, on the perpendicular bisector of the segment $O_{1} O_{2}$).

![](https://cdn.mathpix.com/cropped/2024_05_02_d7e0f62eafce6a75fbfeg-1.jpg?height=558&width=829&top_left_y=1244&top_left_x=629)

Figure 2

Based on this, the construction of the desired point is straightforward: the intersection points of the Apollonian circles corresponding to the points $O_{1}$ and $O_{2}$ and the ratio $r_{1}: r_{2}$, and the points $O_{2}$ and $O_{3}$ and the ratio $r_{2}: r_{3}$ will be the desired points (for these points $R$, $\frac{O_{1} R}{O_{2} R} = \frac{r_{1}}{r_{2}}$ and $\frac{O_{2} R}{O_{3} R} = \frac{r_{2}}{r_{3}}$ imply $\frac{O_{1} R}{O_{3} R} = \frac{r_{1}}{r_{3}}$, so they also lie on the Apollonian circle corresponding to the points $O_{1}$ and $O_{3}$ and the ratio $r_{1}: r_{3}$).

The number of solutions is 2, 1, or 0, depending on whether the two circles (or the perpendicular bisector) have 2, 1, or 0 common points.