In a right triangle, the altitude from a vertex to the hypotenuse splits the hypotenuse into two segments of lengths $a$ and $b$. If the right triangle has area $T$ and is inscribed in a circle of area $C$, find $ab$ in terms of $T$ and $C$.
1. **Geometric Mean Theorem**: In a right triangle, the altitude from the right angle to the hypotenuse splits the hypotenuse into two segments of lengths \(a\) and \(b\). The length of the altitude \(h\) is given by:
   \[
   h = \sqrt{ab}
   \]

2. **Radius of the Circumscribed Circle**: The hypotenuse of the right triangle is the diameter of the circumscribed circle. Let the hypotenuse be \(c\). Then the radius \(R\) of the circle is:
   \[
   R = \frac{c}{2}
   \]
   The area \(C\) of the circle is:
   \[
   C = \pi R^2 = \pi \left(\frac{c}{2}\right)^2 = \frac{\pi c^2}{4}
   \]

3. **Area of the Triangle**: The area \(T\) of the triangle can be expressed in terms of the altitude \(h\) and the hypotenuse \(c\):
   \[
   T = \frac{1}{2} \times c \times h = \frac{1}{2} \times c \times \sqrt{ab}
   \]

4. **Expressing \(c\) in terms of \(a\) and \(b\)**: Since \(c\) is the hypotenuse, we have:
   \[
   c = a + b
   \]

5. **Substituting \(c\) into the area formula**: Using \(c = a + b\), the area \(T\) becomes:
   \[
   T = \frac{1}{2} \times (a + b) \times \sqrt{ab}
   \]

6. **Expressing \(C\) in terms of \(a\) and \(b\)**: Using \(c = a + b\), the area \(C\) of the circle is:
   \[
   C = \frac{\pi (a + b)^2}{4}
   \]

7. **Relating \(T\) and \(C\)**: We need to find \(ab\) in terms of \(T\) and \(C\). First, we square the area \(T\):
   \[
   T^2 = \left(\frac{1}{2} (a + b) \sqrt{ab}\right)^2 = \frac{1}{4} (a + b)^2 ab
   \]

8. **Dividing \(T^2\) by \(C\)**: We have:
   \[
   \frac{T^2}{C} = \frac{\frac{1}{4} (a + b)^2 ab}{\frac{\pi}{4} (a + b)^2} = \frac{ab}{\pi}
   \]

9. **Solving for \(ab\)**: Multiplying both sides by \(\pi\), we get:
   \[
   ab = \pi \frac{T^2}{C}
   \]

The final answer is \(\boxed{\frac{\pi T^2}{C}}\).