jmc

Since $rs=A$, where $r$ is the inradius, $s$ is the semiperimeter, and $A$ is the area, we have that the ratio of the area of the circle to the area of the triangle is $\frac{\pi r^2}{rs}=\frac{\pi r}{s}$. Now we try to express $s$ as $h$ and $r$. Denote the points where the incircle meets the triangle as $X,Y,Z$, where $O$ is the incenter, and denote $AX=AY=z,BX=BZ=y,CY=CZ=x$. Since $XOZB$ is a square (tangents are perpendicular to radius), $r=BX=BZ=y$. The perimeter can be expressed as $2(x+y+z)$, so the semiperimeter is $x+y+z$. The hypotenuse is $AY+CY=z+x$. Thus we have $s=x+y+z=(z+x)+y=h+r$. The answer is $\boxed{\frac{\pi r}{h+r}}$.