2 circles are tangent with each other at one point. One has a radius of 12 with a center named P and another has a radius of 8 with a center named Q. A common external tangent intercepts line PQ at R. Find QR.

Asymptote code below
[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]

To solve the problem of finding $ QR $, we can use geometry principles involving circles, tangents, and right triangles.

### Step-by-Step Explanation:

We have two circles:
- Circle 1 with radius $ r_1 = 12 $ and center $ P $,
- Circle 2 with radius $ r_2 = 8 $ and center $ Q $.

The two circles are tangent to each other externally at one point, and a common external tangent intersects the line $ PQ $ at point $ R $. We need to find the distance $ QR $.

### Geometry Insights:

1. **Distance Between Centers (PQ):**
   Since the circles are externally tangent, the distance between their centers $ P $ and $ Q $ is the sum of their radii:
   \[
   PQ = r_1 + r_2 = 12 + 8 = 20
   \]
   
2. **Right Triangle Setup:**
   The common external tangent touches both circles at a point, and the tangent at each circle forms a right angle with the radius at the point of tangency. This means we can form a right triangle where:
   - The hypotenuse is $ PQ = 20 $,
   - The vertical legs are the radii of the circles: $ r_1 = 12 $ and $ r_2 = 8 $,
   - The segment $ QR $ is parallel to the common external tangent.

3. **Use of the Tangent Length Formula:**
   The length of the common external tangent between two circles can be calculated using the following formula:
   \[
   \text{Length of tangent} = \sqrt{PQ^2 - (r_1 - r_2)^2}
   \]
   Plugging in the values:
   \[
   \text{Length of tangent} = \sqrt{20^2 - (12 - 8)^2} = \sqrt{400 - 16} = \sqrt{384} = 8\sqrt{6}
   \]
   
   Therefore, the length of the tangent (segment $ QR $) is $ 8\sqrt{6} $.

### Conclusion:
The length of $ QR $ is $ 8\sqrt{6} $.

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Let me know if you'd like more details or have any questions! 

Here are five related questions:
1. How do you find the distance between two tangent circles?
2. What is the formula for the length of a common internal tangent?
3. How does the geometry of tangents change if the circles are internally tangent?
4. How do you prove that the tangent to a circle is perpendicular to the radius at the point of tangency?
5. What other geometric problems can be solved using similar right-triangle setups?

**Tip:** When dealing with tangent lines and circles, remember that the tangent is always perpendicular to the radius at the point of contact.

2 circles are tangent with each other at one point. One has a radius of 12 with a center named P and another has a radius of 8 with a center named Q. A common external tangent intercepts line PQ at R. Find QR.

Explore how to find the length of QR in a problem involving two tangent circles with radii 12 and 8, a common external tangent, and line PQ. This solution uses geometry principles like the tangent-secant theorem and Pythagorean Theorem. Suitable for Grades 10-12.

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