Finding out an arc's radius by arc length and endpoints

I have two points. I need to draw an arc ($<180$°) between them, and I know how long it should be, but nothing else about it.

Knowing either the radius length or the coordinates of the center point of the circle should be enough to draw it directly. I could also calculate the radius if I knew the subtended angle θ. But I don't know how to calculate either of these.

I know that somebody answered a similar question a while ago. But for some reason, the formula they give for the answer includes θ. Maybe there is an easy way to calculate it and my knowledge is just too rusty, but I can't solve my problem with this information.

I tried deriving an answer by myself, but arrived at an equation containing both an angle's cosine and the angle squared, and I don't know how to solve such an equation. Besides, there is probably an easier way than what I used.

It is OK if the answer is derived by using trigonometry, analytical geometry or anything else, as long as I can use a pocket calculator to get my radius/coordinates. But if you don't want to lose me along the way, it would be nice if you could use calculations at undergraduate level.

Here is a drawing of what I mean:

If $r$ is the radius then you have presumably found, as Isaac did, that $l=r\theta$ and $\frac{c}{2}=r \sin(\theta/2).$ You could eliminate $\theta$ to get $$2r \sin\left(\frac{l}{2r}\right) = c.$$ That you can solve for $r$ using numerical methods.

As you said you had, I arrive at an equation that cannot be solved algebraically, $\cos\frac\theta 2=\frac{c\theta}{2l}$, so I suspect that there is no algebraic solution, only a numerical approximation.

The only easier thing comes when the arc length is not much greater than the length of the straight line segment from A to B, which means the angle of the circle is small. Then you can use the small angle approximation $\cos\frac{\theta}{2}=1-\frac{\theta^2}{8}=\frac{c\theta}{2l}$, which can be solved with the quadratic formula: $\theta^2+\frac{4c\theta}{l}-88=0$ $\theta=\frac{2c}{l}+\sqrt{\frac{4c^2}{l^2}+8}$ Of course, when the difference is very small, you can just draw the straight line.