In the figure below, circle $O$ has two tangents,  $\overline{AC}$ and  $\overline{BC}$.  $\overline{EF}$ is drawn tangent to circle $O$ such that $E$ is on  $\overline{AC}$, $F$ is on  $\overline{BC}$, and  $\overline{EF} \perp  \overline{FC}$. Given that the diameter of circle $O$ has length $10$ and that $CO = 13$, what is the area of triangle $EFC$?
[img]https://cdn.artofproblemsolving.com/attachments/b/d/4a1bc818a5e138ae61f1f3d68f6ee5adc1ed6f.png[/img]
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 \(\boxed{\frac{420}{17}}\)