The problem:
We have two circles of radii r1 and r2, respectively, that are tangent to each other externally at point A. PQ is a direct common tangent to both circles. We need to find the value of PQ^2.

Approach:
To solve this problem, we can use the properties of tangents to circles. Let's analyze the given situation and derive the solution step by step.

Step 1: Drawing the diagram:
We begin by drawing the two circles and their common tangent PQ. We also label the centers of the circles as O1 and O2, and the points of tangency as B and C, respectively.

Step 2: Identifying key points:
From the diagram, we can observe the following key points:
- O1, O2: The centers of the circles.
- A: The point of tangency between the two circles.
- B, C: The points of tangency between the circles and the tangent PQ.

Step 3: Identifying key lines and segments:
We also have the following important lines and segments:
- PQ: The direct common tangent to both circles.
- O1A and O2A: Lines connecting the centers of the circles to the point of tangency A.
- OB and OC: Segments connecting the centers of the circles to the points of tangency B and C, respectively.

Step 4: Applying the tangent properties:
Now, let's use the properties of tangents to analyze the given situation and derive the solution.

Property 1: A line drawn from the center of a circle to the point of tangency is perpendicular to the tangent line.
From this property, we can conclude that:
- O1A is perpendicular to PQ at point B.
- O2A is perpendicular to PQ at point C.

Property 2: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
From this property, we can apply it to triangles O1AB and O2AC:
- In triangle O1AB, we have O1A^2 = OB^2 + PQ^2.
- In triangle O2AC, we have O2A^2 = OC^2 + PQ^2.

Step 5: Deriving the solution:
Now, let's substitute the values of O1A and O2A in the above equations:
- O1A = r1 + r2 (sum of the radii of the two circles)
- O2A = r1 - r2 (difference of the radii of the two circles)

Substituting these values in the above equations, we get:
- (r1 + r2)^2 = OB^2 + PQ^2
- (r1 - r2)^2 = OC^2 + PQ^2

Since OB = OC (both are radii of the same circle), we can equate the right-hand sides of the above two equations:
OB^2 + PQ^2 = OC^2 + PQ^2

Simplifying the equation, we get:
(r1 + r2)^2 = (r1 - r2)^2

Expanding both sides of the equation, we get:
r1