How do I fit two equal-length arcs in the opposite corners of a right triangle? I am trying to make two arcs of the same length such that they will both fit in a right triangle like so:

I am given the first radius (r1) and a constant k that is the difference between the leg opposite angle one and the second radius (r2)
I know that the angle of the arcs have to add up to Pi/2. I also know that this becomes impossible at some small value of k - for instance, if k was zero there is no way the two arcs could be of equal length.
If I knew the angle, I could find the second radius, or vice-versa, using trigonometric functions.
In the case where both the angles were Pi/4, I know that k would be r*(sqrt(2) - 1).

In case anyone wants to know why I'm doing this, I'm trying to show a circle that is broken open at a point and the broken ends curve out until they are parallel and a certain distance apart. That distance will be 2*k. I want to make sure that the curved-outward parts are the same length as they would be if they were still completing the circle.

just hint
The same length means that
$$r_1\theta =r_2 (\frac {\pi}{2}-\theta) $$
or
$$\theta=\frac {\pi r_2}{2 (r_1+r_2)} $$
in the triangle, 
$$\sin (\theta)=\frac {k+r_2}{r_1+r_2} $$
if we put $k=xr_2$ then
$$x=\frac {2\theta}{\pi}\sin (\theta)-1$$
