1. **Problem statement:** We have two towers represented by two circles with radii 100 km and 150 km. A common external tangent line CE touches the circles at points C and E. We want to find the distance between the centers of the two towers. 2. **Known values:** - Radius of tower A (circle A): $r_1 = 100$ km - Radius of tower B (circle B): $r_2 = 150$ km - Tangent segment CE touches circle A at C and circle B at E 3. **Key concept:** For two circles with centers separated by distance $d$, and radii $r_1$ and $r_2$, the length of the common external tangent segment $CE$ satisfies the relation: $$CE^2 = d^2 - (r_1 + r_2)^2$$ This comes from the right triangle formed by the centers and the tangent points. 4. **Given:** The problem states segment CE is the common tangent, but does not give its length directly. However, it provides segments CD = 200 km and DE = 300 km, so: $$CE = CD + DE = 200 + 300 = 500 \text{ km}$$ 5. **Apply the formula:** $$CE^2 = d^2 - (r_1 + r_2)^2$$ Substitute known values: $$500^2 = d^2 - (100 + 150)^2$$ $$250000 = d^2 - 250^2$$ $$250000 = d^2 - 62500$$ 6. **Solve for $d^2$:** $$d^2 = 250000 + 62500 = 312500$$ 7. **Find $d$:** $$d = \sqrt{312500} = \sqrt{3125 \times 100} = 10 \sqrt{3125}$$ Since $3125 = 25 \times 125$, $$d = 10 \times 5 \times \sqrt{125} = 50 \sqrt{125}$$ And $\sqrt{125} = \sqrt{25 \times 5} = 5 \sqrt{5}$, so $$d = 50 \times 5 \sqrt{5} = 250 \sqrt{5}$$ 8. **Final answer:** The distance between the two towers is: $$\boxed{250 \sqrt{5} \text{ km}}$$ This is approximately $250 \times 2.236 = 559$ km.