5. Circles $a\left(A ; r_{a}\right), b\left(B ; r_{b}\right)$ are given, which touch each other externally at point $T$. Their common external tangent touches circle $a$ at point $T_{a}$ and circle $b$ at point $T_{b}$. Using $r_{a}, r_{b}$, express the ratio of the radii of the circles $k_{a}$, $k_{b}$ circumscribed around the triangles $T_{a} A T$, $T_{b} B T$ respectively.

(Sárka Gergelitsová)
SOLUTION. Let $Q$ be the intersection of the common external tangent $T_{a} T_{b}$ and the common internal tangent of circles $a$ and $b$ at point $T$. A tangent to a circle forms a right angle with the diameter of that circle passing through the point of tangency, so angles $A T_{a} Q$ and $A T Q$ are right angles. Points $T_{a}$ and $T$ thus lie on the circle with diameter $A Q$, which is also the circumcircle $k_{a}$ of triangle $T_{a} A T$ (Fig. 2). Similarly, we can show that $Q B$ is the diameter of circle $k_{b}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_311860a79a0296bd8ddfg-12.jpg?height=562&width=874&top_left_y=787&top_left_x=594)

Fig. 2

Lines $Q T$ and $Q T_{a}$ are tangents to circle $a$, and line $Q A$ passing through its center is thus the angle bisector of $\angle T Q T_{a}$. For similar reasons, line $Q B$ is the angle bisector of $\angle T Q T_{b}$. Angle $T_{a} Q T_{b}$ is a right angle, so lines $A Q$ and $B Q$ form a right angle, and triangle $A B Q$ is a right triangle. The length of its leg $A Q$ is given by Eukleid's theorem as $|A Q|=\sqrt{r_{a}\left(r_{a}+r_{b}\right)}$ and for the length of the other leg we have $|B Q|=\sqrt{r_{b}\left(r_{a}+r_{b}\right)}$.

The desired ratio of the radii of the circumcircles of triangles $T_{a} A T$ and $T_{b} B T$ is equal to the ratio of the lengths of their diameters $A Q$ and $B Q$, i.e.,

$$
\frac{|A Q|}{|B Q|}=\frac{\sqrt{r_{a}\left(r_{a}+r_{b}\right)}}{\sqrt{r_{b}\left(r_{a}+r_{b}\right)}}=\sqrt{\frac{r_{a}}{r_{b}}}
$$