A circle with center $O$ has radius $8$ units and circle $P$ has radius $2$ units.  The circles are externally tangent to each other at point $Q$.  Segment $TS$ is the common external tangent to circle $O$ and circle $P$ at points $T$ and $S$, respectively.  What is the length of segment $OS$?  Express your answer in simplest radical form.
We create a diagram with the given information from the problem: [asy]
draw(Circle((0,0),8));
draw(Circle((10,0),2));
dot((0,0));dot((10,0));
label("$O$",(0,0),SW); label("$P$",(10,0),SW);

dot((8,0)); label("$Q$",(8,0),SW);

label("$T$",(4.6,6.6),NE); label("$S$",(11,1.7),NE);
draw((4.6,6.6)--(11,1.7));
[/asy]

We draw in radii $OT$ and $PS$ and connect $O$ and $P$.  Then we drop a perpendicular from $P$ to $OT$ that intersects $OT$ at $R$:

[asy]
draw((0,0)--(4.6,6.6),red); draw((10,0)--(11,1.7),blue);
draw(Circle((0,0),8));
draw(Circle((10,0),2));
dot((0,0));dot((10,0));
label("$O$",(0,0),SW); label("$P$",(10,0),SW);

label("$T$",(4.6,6.6),NE); label("$S$",(11,1.7),NE);
draw((4.6,6.6)--(11,1.7));
draw((0,0)--(8,0),red); draw((8,0)--(10,0),blue);
draw((10,0)--(3.3,4.8));
label("$R$",(3.3,4.8),W);
[/asy]

$\angle OTS$ and $\angle PST$ are right angles as tangents create right angles with radii at points of tangency.  $RTSP$ is a rectangle, and $\triangle ORP$ is right.  We use Pythagorean theorem on $\triangle ORP$:  we have $OP=8+2=10$ and $OR=8-2=6$, so $RP=\sqrt{OP^2-OR^2}=\sqrt{10^2-6^2}=8$.  Then $TS=8$ as well.

[asy]
draw((0,0)--(4.6,6.6));

label("8",(2,3),N); label("8",(8,5));
draw(Circle((0,0),8));
draw(Circle((10,0),2));
dot((0,0));dot((10,0));
label("$O$",(0,0),SW); label("$P$",(10,0),SW);

label("$T$",(4.6,6.6),NE); label("$S$",(11,1.7),NE);
draw((4.6,6.6)--(11,1.7));
draw((0,0)--(11,1.7));
draw((10,0)--(11,1.7));
[/asy] Finally, $OS$ is the hypotenuse of right triangle $\triangle OTS$ with $OT=TS=8$.  Hence $OS=\sqrt{8^2+8^2}=\boxed{8\sqrt{2}}$.
8\sqrt{2}
Level 5
Geometry
Problem: A circle with center $O$ has radius $8$ units and circle $P$ has radius $2$ units.  The circles are externally tangent to each other at point $Q$.  Segment $TS$ is the common external tangent to circle $O$ and circle $P$ at points $T$ and $S$, respectively.  What is the length of segment $OS$?  Express your answer in simplest radical form.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the length of segment OS, we first need to understand the configuration of the circles and the tangent line. Circle O has a radius of 8 units, and circle P has a radius of 2 units. The distance between the centers of the two circles, denoted as OP, is the sum of their radii, which is 8 + 2 = 10 units.

Next, we apply the formula for the length of the common external tangent between two circles. The formula is given by:

L = √(d² - (r1 - r2)²)

where L is the length of the tangent, d is the distance between the centers of the circles, r1 is the radius of the larger circle, and r2 is the radius of the smaller circle.

In our case, d = 10, r1 = 8, and r2 = 2. We first calculate the difference in radii:

r1 - r2 = 8 - 2 = 6.

Now we square this difference:

(r1 - r2)² = 6² = 36.

Next, we square the distance between the centers:

d² = 10² = 100.

Now we substitute these values into the tangent length formula:

L = √(100 - 36) = √(64) = 8.

The length of the tangent TS is 8 units. 

To find OS, we recognize that OS is the distance from the center of circle O to the point of tangency S on circle P. Since TS is a tangent, it forms a right triangle with the radius of circle P at point S. The radius of circle P is 2 units, and the length of TS is 8 units.

Using the Pythagorean theorem in triangle OST, where OT is the radius of circle O (8 units), OS is the hypotenuse, and TS is the tangent (8 units), we can find OS:

OS² = OT² + TS²
OS² = 8² + 8²
OS² = 64 + 64
OS² = 128.

Taking the square root gives us:

OS = √(128) = √(64 * 2) = 8√2.

Thus, the length of segment OS is 8√2.<|END_OF_LATENT|> The final answer is $8\sqrt{2}$.