The length of the hypotenuse of a right triangle is $h$ , and the radius of the inscribed circle is $r$. The ratio of the area of the
circle to the area of the triangle is

$\textbf{(A) }\frac{\pi r}{h+2r}\qquad\textbf{(B) }\frac{\pi r}{h+r}\qquad\textbf{(C) }\frac{\pi}{2h+r}\qquad\textbf{(D) }\frac{\pi r^2}{r^2+h^2}\qquad\textbf{(E) }\text{none of these}$
1. Let the sides of the right triangle be \(a\), \(b\), and \(c\) where \(c\) is the hypotenuse. Given that \(c = h\).
2. The area \(A\) of the triangle is given by:
   \[
   A = \frac{1}{2}ab
   \]
3. The semi-perimeter \(S\) of the triangle is:
   \[
   S = \frac{a + b + c}{2}
   \]
4. The radius \(r\) of the inscribed circle is related to the area and the semi-perimeter by the formula:
   \[
   A = rS
   \]
5. Substituting the semi-perimeter \(S\) into the formula for the area:
   \[
   \frac{1}{2}ab = r \left( \frac{a + b + c}{2} \right)
   \]
6. Simplifying, we get:
   \[
   ab = r(a + b + c)
   \]
7. Since \(c = h\), we have:
   \[
   ab = r(a + b + h)
   \]
8. The area of the inscribed circle is:
   \[
   \text{Area of the circle} = \pi r^2
   \]
9. The ratio of the area of the circle to the area of the triangle is:
   \[
   \frac{\text{Area of the circle}}{\text{Area of the triangle}} = \frac{\pi r^2}{\frac{1}{2}ab}
   \]
10. From step 6, we know that \(ab = r(a + b + h)\), so:
    \[
    \frac{\pi r^2}{\frac{1}{2}ab} = \frac{\pi r^2}{\frac{1}{2}r(a + b + h)} = \frac{\pi r^2}{\frac{r(a + b + h)}{2}} = \frac{\pi r^2 \cdot 2}{r(a + b + h)} = \frac{2\pi r^2}{r(a + b + h)} = \frac{2\pi r}{a + b + h}
    \]
11. Since \(a + b + h = 2S\), we can rewrite the ratio as:
    \[
    \frac{2\pi r}{2S} = \frac{\pi r}{S}
    \]
12. Given that \(S = \frac{a + b + h}{2}\), we have:
    \[
    \frac{\pi r}{S} = \frac{\pi r}{\frac{a + b + h}{2}} = \frac{2\pi r}{a + b + h}
    \]
13. Since \(a + b + h = 2r + h\), we get:
    \[
    \frac{2\pi r}{2r + h}
    \]

The final answer is \(\boxed{\frac{\pi r}{h + 2r}}\)