7. Circles $k_{1}$ and $k_{2}$ touch each other externally. The ratio of the lengths of their radii is $2: 1$. A common tangent touches circle $k_{1}$ at point $T$ and circle $k_{2}$ at point $R$. What is the ratio of the length of segment $\overline{T R}$ to the length of the radius of the larger circle?
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_a89c08f98accc4763978g-4.jpg?height=665&width=974&top_left_y=1592&top_left_x=541)

Sketch:

$1 \text{ POINT}$

Let $S_{1}$ and $S_{2}$ be the centers of circles $k_{1}$ and $k_{2}$, respectively, and their radii $r_{1}$ and $r_{2}$. Assume $r_{1} > r_{2}$. The problem is solved similarly in the case $r_{2} > r_{1}$. Draw a line $x$ parallel to $\overline{T R}$ through point $S_{2}$. Let $P$ be the intersection of line $x$ and the radius $\overline{S_{1} T}$ of circle $k_{1}$.

Then, triangle $S_{1} S_{2} P$ is a right triangle, 1 POINT and quadrilateral $S_{2} P T R$ is a rectangle, with $\left|S_{2} P\right| = |T R|$.

By applying the Pythagorean theorem, we get $|T R| = \left|P S_{2}\right| = \sqrt{\left|S_{1} S_{2}\right|^{2} - \left|S_{1} P\right|^{2}}$.

2 POINTS

Use the ratio (2:1) between the radii of circles $k_{1}$ and $k_{2}$ and the assumption that $r_{1} > r_{2}$. There exists a number $r$ such that we can write $r_{1} = 2r$, $r_{2} = r$.

Then, $\left|S_{1} S_{2}\right| = 3r$.

Substituting into $\left({ }^{*}\right)$, we get $|T R| = \sqrt{9r^{2} - r^{2}} = \sqrt{8r^{2}} = 2\sqrt{2}r$.

Thus, the desired ratio is $|T R| : \left|S_{1} T\right| = 2\sqrt{2}r : 2r = \sqrt{2} : 1$.

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