1. **Problem Statement:** We have a circle centered at $O$ with tangent segments $\overline{EH}$ and $\overline{EF}$ from point $E$ to the circle. The segment $\overline{HF}$ passes through the center $O$. Given $EH = 15.3$ and $HF = 7.2$, we need to find the length of $HG$. 2. **Key Concepts:** - Tangent segments from a common external point to a circle are equal in length. So, $EH = EF = 15.3$. - Since $HF$ passes through the center $O$, $HF$ is a diameter or part of a diameter. - Point $G$ lies on the circle between $H$ and $F$, so $HG + GF = HF$. 3. **Step-by-step Solution:** - Since $HF$ passes through the center $O$, $O$ lies on $HF$. - $G$ is the point of tangency on the circle along $HF$, so $OG$ is the radius. - Because $EH$ and $EF$ are tangents from $E$, $EH = EF = 15.3$. - The radius $OG$ is perpendicular to the tangent at $G$. - Since $HF$ passes through $O$ and $G$ lies on the circle, $HG$ is the radius segment from $H$ to $G$. - Given $HF = 7.2$, and $HG$ is the radius, then $GF = HF - HG$. 4. **Using the tangent-secant theorem:** - The power of point $E$ with respect to the circle is $EH^2 = EF^2 = EG \times EH$ (where $EG$ is the secant segment). - But here, since $EH$ and $EF$ are tangents, $EH^2 = EG \times EF$ is not directly applicable. 5. **Using right triangle properties:** - Since $OG$ is radius and $HG$ is radius, and $HF$ is a chord passing through $O$, $HG$ is the radius. - The length $HG$ is the radius of the circle. 6. **Calculate $HG$:** - Since $HF = 7.2$ and $O$ is the center, $HF$ is a diameter or part of it. - If $HF$ is the diameter, then radius $HG = \frac{HF}{2} = \frac{7.2}{2} = 3.6$. **Final answer:** $$HG = 3.6$$