1. Identify the given information and draw the necessary elements:
- Circle $O$ has a diameter of 10, so the radius $r = 5$.
- $\overline{AC}$ and $\overline{BC}$ are tangents to the circle at points $A$ and $B$ respectively.
- $\overline{EF}$ is tangent to the circle at point $D$ and perpendicular to $\overline{FC}$.
- $CO = 13$.

2. Determine the lengths of segments using the Pythagorean theorem:
- Since $\overline{AC}$ and $\overline{BC}$ are tangents from a common external point $C$, $\overline{CA} = \overline{CB}$.
- Let $OB = 5$ (radius), and since $\angle OBF = 90^\circ$, $OBFD$ forms a square with side length 5.
- Therefore, $BF = FD = 5$.

3. **Calculate $BC$ using the Pythagorean theorem:**
$$BC = \sqrt{OC^2 - OB^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$$

4. **Determine $FC$:**
$$FC = BC - BF = 12 - 5 = 7$$

5. **Set up the equation for $\overline{EF}$:**
- Let $AE = x$.
- Then $ED = x$ and $EC = 12 - x$.
- $FE = ED + DF = x + 5$.

6. **Apply the Pythagorean theorem to triangle $EFC$:**
$$FC^2 + FE^2 = EC^2$$
$$7^2 + (x + 5)^2 = (12 - x)^2$$
$$49 + x^2 + 10x + 25 = 144 - 24x + x^2$$
$$74 + 10x = 144 - 24x$$
$$34x = 70$$
$$x = \frac{70}{34} = \frac{35}{17}$$

7. **Calculate $EF$:**
$$EF = x + 5 = \frac{35}{17} + 5 = \frac{35}{17} + \frac{85}{17} = \frac{120}{17}$$

8. **Find the area of triangle $EFC$:**
$$\text{Area} = \frac{1}{2} \times EF \times FC = \frac{1}{2} \times \frac{120}{17} \times 7 = \frac{1}{2} \times \frac{840}{17} = \frac{420}{17}$$

The final answer is $\boxed{\frac{420}{17}}$