Task 66. Find the distance from the point of tangency of two circles with radii $R$ and $r$ to their common external tangent.

## Java 9. Lemma about the perpendicular from the point of tangency

## 

Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. 

---

Task 66. Find the distance from the point of tangency of two circles with radii $R$ and $r$ to their common external tangent.

## Java 9. Lemma about the perpendicular from the point of tangency
Solution. Let two circles with centers $O_{1}$ and $O_{2}$ touch each other at point $C$ and have a common external tangent at points $A$ and $B$. The problem requires finding the distance $C D$ (Fig. 5). The angle $A C B$ is a right angle (see problem 83), so if we construct a circle with diameter $A B$, it will touch the line of centers and the radii $O_{1} A$ and $O_{2} B$. Then, by Archimedes' lemma, the line $O_{1} B$ bisects the perpendicular $C D$ and the line $O_{2} A$ bisects $C D$, meaning that $O_{1} B$ and $O_{2} A$ intersect on the segment $C D$. According to a known problem, the length of the segment parallel to the bases of a trapezoid through the intersection of the diagonals is equal to the harmonic mean of the lengths of the bases. Therefore, in trapezoid $O_{1} A B O_{2}$, the segment $C D$ is equal to $\frac{2 R r}{R+r}$.

## Chapter 10. THE LEMMA ABOUT THE CENTER OF A CIRCLE

One of the problems in "The Book of Lemmas" is formulated as follows:

A chord $A C$ cuts off a segment from a circle. On the arc bounding this segment, points $B$ and $F$ are taken such that $A B = B F$ (Fig. 56). A perpendicular $B D$ is dropped from point $B$ to the chord $A C$, and a point $E$ is taken on $A C$ such that $A D = D E$. Then $C E = C F$.

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-049.jpg?height=522&width=557&top_left_y=704&top_left_x=1155)

It is not difficult to notice the connection between this lemma and one of the modern problems:

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-049.jpg?height=622&width=576&top_left_y=1328&top_left_x=363)

Two circles intersect at points $A$ and $F$, and one of the circles passes through the center $B$ of the other. A point $C$ is taken on the first circle (Fig. 57). The segment $A C$ intersects the second circle at point $E$. Prove that $E C = C F$.

As we can see, in the given condition, instead of three equal segments $A B, B E$, and $B F$ (as in Fig. 56), a circle with its center on the given circle is introduced.

We will use the modern formulation, as it is less cumbersome. This is the problem we will call the "LEMMA ABOUT THE CENTER OF A CIRCLE."

Consider two methods of proving this lemma.

48

Proof 1. So, we need to prove that $C E = C F$ (see Fig. 57). The quadrilateral $A B F C$ is cyclic, so

$$
\angle B F C + \angle B A C = 180^{\circ}
$$

But since $\angle B A E = \angle B E A$ (triangle $B A E$ is isosceles), then

$$
\angle B F C + \angle B E A = 180^{\circ}
$$

which means $\angle B F C = \angle B E C$. Considering that $\angle B F E = \angle B E F$, we get: $\angle C F E = \angle C E F$, from which the required equality follows.

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-050.jpg?height=617&width=648&top_left_y=754&top_left_x=356)

Fig. 58

Proof 2. Using the notation of Fig. 58, we need to prove that $C D = C B$. Let $\angle D A B = \alpha$. Then $\angle D O_{2} B = 2 \alpha$. But

$$
\angle D A B = \angle C A B = \angle C O_{2} B
$$

since they subtend the same arc $B C$. Therefore, $\angle C O_{2} B = \alpha$, and thus $\angle D O_{2} C = \alpha$ and

$$
\Delta C O_{2} D = \Delta C O_{2} B
$$

by two sides and the included angle. Hence, $C D = C B$.

Note that the lemma is also valid for other cases of the mutual arrangement of the circles and their points of intersection.

Let's prove that $C D = D B$ if the circles are arranged as shown in Fig. 59.

Let $\angle B C D = \alpha$, then $\angle B O_{2} A = 2 \alpha$, from which $\angle B D A$ is also equal to $2 \alpha$ (subtends the same arc). In $\triangle B D C$, $\angle B D A$ is an exterior angle. It is easy to see that $\angle C B D$ is also equal to $\alpha$. Thus, we get that triangle $B D C$ is isosceles.

Consider the application of the lemma about the center of a circle.

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-051.jpg?height=586&width=602&top_left_y=398&top_left_x=316)