Distance between the nail and the center of the disk 
Suppose you have a disk with radius $r$ and a string of length $2
\pi r+l$, i.e. longer than the perimeter of the disk. Hang the disk (of center $O$) from the nail at $A$ using the string as shown in the picture:



Then what the distance $OA$ equals in terms of $r$ and $l$ ?

At first, this problem seemed so straightforward that I didn't try to solve. But when I try headaches came over. After a couple of hours with failing some basic geometric attempts, I started to find a differential equation that $OA$ as function of $l$ will satisfy, but without success. So, I have no any other clue for solving this problem, except sharing it with community.
 A: I've labeled points $B$ and $C$ and angles $a$ and $b$ (in radians) in this diagram:

The arc subtended by the angle $b$ is just $br$, and this lets us find the length of $\overline{AB}$ and $\overline{AC}$ as
$$br + \frac{l}{2}$$
Now we can use the Pythagorean theorem to find the length of $\overline{AO}$ in terms of $b$, $l$, and $r$:
$$\sqrt{\left(br + \frac{L}{2}\right)^2 + r^2}=\sqrt{b^2r^2+r^2+brl+\frac{l^2}{4}}$$

We still have to find the angle $b$, and as others have already pointed out, this isn't easy. Notice that
$$\tan b = \frac{2br+l}{2r} \qquad\Rightarrow\qquad b = \arctan{\frac{2br+l}{2r}}$$
If you have to do this by hand, then an approximation using Newton's method (as suggested by another user) is probably appropriate. I would have suggested the Taylor series for $\arctan$ but that only converges for arguments with absolute value less than $1$. 
A: If the distance between the nail and the center of the disk is $d$, then the length of the string is given by:
$$ 2\sqrt{d^2-r^2}+2\left(\pi-\arccos\frac{r}{d}\right),$$
so, in order to find $d$ given $l$ and $r$ you have to solve a trascendental equation. 
Newton's method is well-suited for such a task.
A: Let $T$ one of the tangency points, $\alpha=\angle OAT$, $x=OA$. Note that $AT$ and $OT$ are perpendicular.
Then
$$2\pi r+l=2[(\pi-\alpha)r+\sqrt{x^2-r^2}]$$
but $\alpha=\arccos(x/r)$. We see that $x$ are inside a square root and inside an $\arccos$ in the same equation. I don't think that this equation can be solvable by algebraic means.
