Mathematics and the art of linearizing the circle

What you want can be achieved using circle arcs, centered at $(0,r)$, of radius $r$ and central angle $2\pi R/r$, with $r$ varying between $1$ and $+\infty$. But I don't know if that is "natural" or not. Here's how it looks:

EDIT.

Each endpoint of the arc describes a spiral curve, given by: $$ x={\pi R\over 2\theta}\sin{2\theta},\quad y={\pi R\over 2\theta}\left(1-\cos{2\theta}\right), $$ where $\theta$ is the polar angle. This also gives the simple polar equation: $$ \rho={\pi R}{\sin \theta\over\theta},\quad 0\le\theta\le{\pi\over2}. $$

If we want this transition to have all the points on the boundary of a circle at all times, then it makes most sense to parameterize by the radius of this circle (and apply a transformation to get it in terms of finite time later). For simplicity, I will also have the transition be to a vertical line.

We shall have the radius of the circle $C_r$ be $r$ and centre be $(-r,0)$, such that $(0,0)$ is on $C_r$ for all $r$. The coordinates of the point at arclength $s$ from $(0,0)$ are then given by $(r (\cos(s/r) - 1), r \sin(s/r))$.

The most natural way to transition would likely be varying the curvature at a constant rate; thus we create a family of curves $f_t:[-\pi, \pi]\to\mathbb{R}^2$ where $t\in[0,1]$ by \begin{align} f_t(s) &= \left(\frac{\cos(s(1-t))-1}{1-t}, \frac{\sin(s(1-t))}{1-t}\right)&t<1\\ f_1(s) &= (0, s)& \end{align}

Since the goals here seem to be rather subjective, I would attempt this and see how it looks to you (beyond making substitutions as needed to result in a horizontal line).