Answer:
2r (1 + √3)
Explanation:
Circle O₁ is tangent to AB. Let's call the point of intersection point D. If we draw a radius from the center O₁ to D, we know this forms a right angle.
△ABC is an equilateral triangle, so we know m∠A = 60°. If we draw a line from A to O₁, we know that bisects the angle, so m∠DAO₁ = 30°.
So △DAO₁ is a 30-60-90 triangle. We can find the length AD:
AD = r √3
Now on the other side, circle O₃ is tangent to AB. Let's call the point of intersection point E. We know it's the same triangle we found earlier, so:
EB = r √3
And finally, we can draw a rectangle connecting O₁, O₃, E, and D. The distance between O₁ and O₃ is 2r, so:
DE = 2r.
Therefore:
AB = r√3 + 2r + r√3
AB = 2r√3 + 2r
AB = 2r (1 + √3)
Here's a graph showing the steps. Hopefully this helps, let me know if you have questions!
desmos.com/calculator/hgaonfzxsm