Bending a line segment REVISED QUESTION
With the help of the existing answers I have been able to put together this clearer animation, and I asked this question to discover the shape is called a cochleoid.  What I am really trying to find out at this point is the following:
It seems like this curve should be perfectly smooth through the point at $(0, 1)$, but because the curve is based on $sin(x)/x$ it is technically undefined at this point, and if I were programming a function to evaluate this curve I would have to add a special case for values near this singularity.  I'm curious if there is any way to re-phrase this equation to remove the singularity and make evaluation of it near zero more numerically stable.
--- Original text of question preserved below ----------------------------------------------------
Imagine I have a unit line segment going from $(0, 0)$ to $(0, 1)$.  Over time, I want to bend this segment such that it always forms a circular arc (the initial configuration can be considered an arc on a circle with infinite radius).  What shape will the end point trace out, and how can I get the coordinates of that point if I am given as input the angle that should be spanned by the arc length?
Edit: Thanks to John Bales below, this animation accurately depicts what I am trying to describe, although I'm not sure how to render the bent segment itself, which would always connect back to the origin.  Is this shape a cartioid?  Is there a way to rephrase the equation so it doesn't become undefined when the input is zero?
 A: The angle in radians subtended by an arc of length $s$ on the circumference of a circle of radius $r$ is given by
\begin{equation}
\theta=\dfrac{s}{r}
\end{equation}
In this instance $s=1$ and $r\ge\tfrac{1}{\pi}$. The circle has center $(r,0)$ and radius $r$. The arc $s$ of length $1$ extends upward along the circumference with one end fixed at $(0,0)$ and the other end ends at the point 
\begin{equation}
(x,y)=\left(r-r\cos\left(\frac{1}{r}\right),r\sin\left(\frac{1}{r}\right)\right)
\end{equation}
These are parametric equations of the curve.
Here is a desmos.com animation of the curve along which the point moves.
https://www.desmos.com/calculator/row6dlgqom
This curve has the following polar equation:
\begin{equation}
r=\dfrac{\sin\left(\frac{\pi}{2}-\theta\right)}{\frac{\pi}{2}-\theta}
\end{equation}
Here is a desmos.com graph of the polar curve with the particle moving along it. You want just the portion of the curve in the first quadrant.
https://www.desmos.com/calculator/rbusr85zsc
I think it is interesting that this turns out to be a $\dfrac{\sin x}{x}$ curve but in the polar coordinate system.
If instead of the interval from $(0,0)$ to $(0,1)$ we bend the unit interval on the horizontal axis along circles with centers $\left(0,\frac{1}{2\theta}\right)$ and radii $\frac{1}{2\theta}$ as $0\le\theta\le\frac{\pi}{2}$ then the ends of the unit arcs along the circles from the origin will trace out the portion of the polar curve
\begin{equation}
r=\dfrac{\sin\theta}{\theta}
\end{equation}
in quadrant I.
The following is a link to a GeoGebra animation illustrating this variation of the problem.
https://www.geogebra.org/m/THDbt3Ad

