The secant passing through the intersection points of the circle with center $O_{1}$ and radius $4 \mathrm{~cm}$ and the circle with center $O_{2}$ and radius $6 \mathrm{~cm}$ intersects the segment $O_{1} O_{2}$ at point $T$, and we know that the length of this segment is not less than $6 \mathrm{~cm}$. The larger circle intersects the segment $O_{1} O_{2}$ at point $A$, and the smaller circle intersects it at point $B$, and $A T: B T=1: 2$. Calculate the length of the segment $O_{1} O_{2}$.
I. Solution. Using the notations of Figure 1. Let $A T=x$ (thus $B T=2 x$) and $T C=y$. The triangles $O_{1} T C$ and $\mathrm{O}_{2} \mathrm{TC}$ are right-angled, and we can apply the Pythagorean theorem to their sides:

$$
\begin{aligned}
(4-2 x)^{2}+y^{2}=4^{2}, & (6-x)^{2}+y^{2}=6^{2} \\
16-16 x+4 x^{2}+y^{2}=16, & 36-12 x+x^{2}+y^{2}=36 \\
4 x^{2}+y^{2}=16 x, & x^{2}+y^{2}=12 x
\end{aligned}
$$

From this, $4 x^{2}+y^{2}=x^{2}+y^{2}+4 x, 3 x^{2}=4 x$. (Here, if $x=0$, the circles would touch each other instead of intersecting, so $x \neq 0$.) Therefore, $x=\frac{4}{3}$, from which

$$
O_{1} O_{2}=4+6-3 x=6 \quad(\mathrm{~cm})
$$

The distance between the centers of the circles is $6 \mathrm{~cm}$ (which means that the center of the smaller circle lies exactly on the larger circle).

![](https://cdn.mathpix.com/cropped/2024_05_02_3623d419fc67596b1577g-1.jpg?height=444&width=599&top_left_y=800&top_left_x=752)

Figure 1

II. Solution. Using the notations of the first solution (Figure 2). Let the other intersection point of the line $O_{1} O_{2}$ with the circle centered at $O_{2}$ be $A^{\prime}$, and the other intersection point with the circle centered at $O_{1}$ be $B^{\prime}$. By Thales' theorem, the triangles $B C B^{\prime}$ and $A^{\prime} C A$ are right-angled, and $C T$ is the altitude to the hypotenuse in both. Applying the altitude theorem to them:

$$
B^{\prime} T \cdot B T=T C^{2} \quad \text { and } \quad A^{\prime} T \cdot A T=T C^{2}, \quad \text { so } \quad B^{\prime} T \cdot B T=A^{\prime} T \cdot A T
$$

$(8-2 x) \cdot 2 x=(12-x) \cdot x$, where $x \neq 0$, otherwise the circles would touch each other. Thus, $(8-2 x) \cdot 2=12-x$, from which $x=\frac{4}{3}$, so $O_{1} O_{2}=6(\mathrm{~cm})$.

![](https://cdn.mathpix.com/cropped/2024_05_02_3623d419fc67596b1577g-1.jpg?height=406&width=629&top_left_y=1781&top_left_x=737)

Figure 2