1. **Problem statement:** We have two circles with centers O and P. Line segment RT is tangent to the larger circle at R and to the smaller circle at S. Given lengths are $OR=9$ m, $PS=3$ m, $ST=6$ m, and $RT=10$ m. We need to find the distance between the centers O and P, rounded to the nearest tenth of a meter. 2. **Understanding the problem:** Since RT is tangent to both circles at points R and S, the radii $OR$ and $PS$ are perpendicular to RT at those points. 3. **Key insight:** The points O, R, S, and P lie such that $OR \perp RT$ and $PS \perp RT$. The distance between centers O and P can be found by considering the right triangles formed by these radii and the tangent line. 4. **Calculate RS:** Since $RT = 10$ m and $ST = 6$ m, then $$RS = RT - ST = 10 - 6 = 4 \text{ m}.$$ 5. **Find distance OP:** The centers O and P lie on a line perpendicular to RT passing through R and S respectively. The distance between O and P is the hypotenuse of a right triangle with legs $OR + PS$ (vertical distance) and $RS$ (horizontal distance). So, $$OP = \sqrt{(OR + PS)^2 + RS^2} = \sqrt{(9 + 3)^2 + 4^2} = \sqrt{12^2 + 4^2} = \sqrt{144 + 16} = \sqrt{160}.$$ 6. **Simplify and round:** $$\sqrt{160} = \sqrt{16 \times 10} = 4\sqrt{10} \approx 4 \times 3.1623 = 12.649.$$ Rounded to the nearest tenth: $$12.6 \text{ m}.$$ **Final answer:** The centers are approximately 12.6 meters apart.