1. **Problem statement:** We have two circles with centers $C$ and $D$ and radii 8 and 10 units respectively. The distance between centers $CD$ is 14 units. The circles intersect at points $F$ and $H$ on a line through $E, C, F, H, D, G$. We need to find the length $HD$. 2. **Key idea:** The points $F$ and $H$ lie on the common chord of the two circles. The segment $CD$ connects the centers. The chord is perpendicular to $CD$ at some point $M$ between $C$ and $D$. We want to find $HD$, the distance from $H$ to $D$ along the line. 3. **Step 1: Find the distance $CM$ from $C$ to the midpoint $M$ of the chord.** Using the intersecting circles formula for chord length and distances: $$CM = \frac{r_C^2 - r_D^2 + CD^2}{2 \times CD}$$ Substitute values: $$CM = \frac{8^2 - 10^2 + 14^2}{2 \times 14} = \frac{64 - 100 + 196}{28} = \frac{160}{28} = \frac{40}{7} \approx 5.714$$ 4. **Step 2: Find $MD$, the distance from $M$ to $D$:** $$MD = CD - CM = 14 - \frac{40}{7} = \frac{98}{7} - \frac{40}{7} = \frac{58}{7} \approx 8.286$$ 5. **Step 3: Find the half-length of the chord $FH = 2 \times HM$ using Pythagoras theorem:** Since $M$ is midpoint of chord $FH$ and perpendicular to $CD$: $$HM = \sqrt{r_D^2 - MD^2} = \sqrt{10^2 - \left(\frac{58}{7}\right)^2} = \sqrt{100 - \frac{3364}{49}} = \sqrt{\frac{4900 - 3364}{49}} = \sqrt{\frac{1536}{49}} = \frac{\sqrt{1536}}{7}$$ Simplify $\sqrt{1536}$: $$1536 = 256 \times 6 \Rightarrow \sqrt{1536} = 16 \sqrt{6}$$ So: $$HM = \frac{16 \sqrt{6}}{7}$$ 6. **Step 4: Find $HD$:** Since $H$ lies on the chord and $M$ is midpoint, $HD = MD - HM$: $$HD = \frac{58}{7} - \frac{16 \sqrt{6}}{7} = \frac{58 - 16 \sqrt{6}}{7}$$ Approximate $\sqrt{6} \approx 2.449$: $$HD \approx \frac{58 - 16 \times 2.449}{7} = \frac{58 - 39.184}{7} = \frac{18.816}{7} \approx 2.688$$ 7. **Step 5: Choose the closest answer:** Options are 2, 4, 6, 8. The closest is 2. **Final answer:** $HD \approx 2$ units.