The two circles below are externally tangent. A common external tangent intersects line $PQ$ at $R.$ Find $QR.$

[asy]
unitsize(0.3 cm);

pair A, B, C, D, E;

A = (0,0);
B = (8,0);
C = (20,0);
D = intersectionpoint(Circle(A,5), arc((A + C)/2, abs(A - C)/2, 0, 180));
E = intersectionpoint(Circle(B,3), arc((B + C)/2, abs(B - C)/2, 0, 180));

draw(Circle(A,5));
draw(Circle(B,3));
draw(A--D);
draw(B--E);

dot("$P$", A, S);
dot("$Q$", B, S);

label("$8$", (B + E)/2, W, red);
label("$12$", (A + D)/2, W, red);
[/asy]

We are given two circles that are externally tangent. A common external tangent intersects the line $ PQ $ at a point $ R $. We are tasked with finding the length of $ QR $, given some geometric information about the circles and the configuration.

Let's break down the problem step by step:

### Step 1: Geometry of the Problem
- Circle 1 is centered at $ P $ with radius 5.
- Circle 2 is centered at $ Q $ with radius 3.
- The distance between the centers $ P $ and $ Q $ is 8 units.
- A common external tangent intersects line $ PQ $ at point $ R $, and we are asked to find the length of $ QR $.

### Step 2: Applying Geometry of Tangents
We use the formula for the distance between the points where the external tangents meet the circles, which is given by:

\[
d = \sqrt{PQ^2 - (r_1 - r_2)^2}
\]

where:
- $ PQ $ is the distance between the centers of the circles (8 units),
- $ r_1 $ is the radius of the first circle (5 units),
- $ r_2 $ is the radius of the second circle (3 units).

### Step 3: Plugging Values into the Formula
Substituting the values into the formula:

\[
d = \sqrt{8^2 - (5 - 3)^2} = \sqrt{64 - 4} = \sqrt{60} = 2\sqrt{15}
\]

This is the distance between the points where the external tangent touches each circle.

### Step 4: Symmetry and Position of Point $ R $
Since $ R $ lies on the line $ PQ $, and the problem's symmetry suggests that the external tangent bisects the line between the two tangent points, we can now find $ QR $ as half of this distance.

Thus, $ QR $ is:

\[
QR = \frac{2\sqrt{15}}{2} = \sqrt{15}
\]

### Final Answer:
The length of $ QR $ is $ \boxed{\sqrt{15}} $.

Would you like more details or have any questions on this problem?

Here are some related questions you can explore:
1. How do we calculate the length of the internal tangent between two circles?
2. What is the general formula for the distance between the external tangents of two circles?
3. How does the point of tangency change if the circles are internally tangent instead of externally tangent?
4. Can we find the equation of the common external tangent for these circles?
5. How would the solution change if the radius of one circle were larger than the other?

**Tip:** When dealing with tangents, symmetry can often simplify calculations by reducing the problem to simpler geometric forms.

The two circles below are externally tangent. A common external tangent intersects line PQ at R. Find QR.

Learn how to calculate the length of QR, where two circles are externally tangent, and a common external tangent intersects line PQ at point R. This problem involves the Pythagorean theorem and properties of tangents. Suitable for students in Grades 9-12.

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