8,9 |

Two circles touch each other externally at point $A$. Find the radii of the circles if the chords connecting point $A$ with the points of tangency from one of the common external tangents are 6 and 8.
Drop a perpendicular from the center of one of the circles to the corresponding chord and consider the similar triangles obtained.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of the circles, $B$ and $C$ be the specified points of tangency ($A B=6, A C=8$). Since triangle $B A C$ is a right triangle (angle $A$ is a right angle), then $B C=10$.

Let $M$ be the foot of the perpendicular dropped from $O_{2}$ to $A C$. From the similarity of triangles $O_{2} M C$ and $C A B$, we find that

$$
O_{2} C=B C \cdot \frac{C M}{A B}=10 \cdot \frac{4}{6}=\frac{20}{3}
$$

Similarly, we find that $O_{1} B=\frac{15}{4}$.

Let $O_{1}$ and $O_{2}$ be the centers of the circles, $B$ and $C$ be the specified points of tangency ($A B=6, A C=8$). Since triangle $B A C$ is a right triangle (angle $A$ is a right angle), then $B C=10$.

Let $M$ be the foot of the perpendicular dropped from $O_{2}$ to $A C$. From the similarity of triangles $O_{2} M C$ and $C A B$, we find that

$$
O_{2} C=B C \cdot \frac{C M}{A B}=10 \cdot \frac{4}{6}=\frac{20}{3}
$$

Similarly, we find that $O_{1} B=\frac{15}{4}$.

Let $O_{1}$ and $O_{2}$ be the centers of the circles, $B$ and $C$ be the specified points of tangency ($A B=6, A C=8$). Since triangle $B A C$ is a right triangle (angle $A$ is a right angle), then $B C=10$.

Let $M$ be the foot of the perpendicular dropped from $O_{2}$ to $A C$. From the similarity of triangles $O_{2} M C$ and $C A B$, we find that

$$
O_{2} C=B C \cdot \frac{C M}{A B}=10 \cdot \frac{4}{6}=\frac{20}{3}
$$

Similarly, we find that $O_{1} B=\frac{15}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-04.jpg?height=426&width=600&top_left_y=267&top_left_x=732)

Answer

$\frac{15}{4} ; \frac{20}{3}$.