A simple geometric problem, solving $f'(x)=\frac{f(x)}{\sqrt{r(x)^2-f(x)^2}}$, given $r(x)$.
Here we assume in general that the plane curve is $C\in\textbf{C}^{(1)}[a,b]$ and of the form
$$
C:\overline{x}(s)=\left\{A(s),B(s)\right\},\tag 1
$$
where $s$ is its canonical parameter i.e.
$$
(A'(s))^2+(B'(s))^2=1.\tag 2
$$
If $P_0\in C$ (a point of the curve) and $t(P_0)$ is the tangent line in $P_0$ and $O_1=(x_1,0)$ is the intersection of $t(P_0)$ with the $x-$axis, then $r(s_0)=|P_0O_1|$. Hence
$$
t(P_0):y-B(s_0)=\frac{B'(s_0)}{A'(s_0)}(x-A(s_0))
$$
and $O_1:y_1=0$. Hence
$$
-B(s_0)=\frac{B'(s_0)}{A'(s_0)}(x_1-A(s_0))\Leftrightarrow x_1-x_0=-B(s_0)\frac{A'(s_0)}{B'(s_0)}.
$$
$$
r(s_0)=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}=\sqrt{B(s_0)^2\frac{(A'(s_0))^2}{(B'(s_0))^2}+(B(s_0)^2)}\Leftrightarrow
$$
$$
r(s_0)=\frac{|B(s_0)|}{|B'(s_0)|}\sqrt{(A'(s_0))^2+(B'(s_0))^2}=\left|\frac{B(s_0)}{B'(s_0)}\right|.
$$
Hence when $s_0$ varies, we get
$$
\frac{B'(s)}{B(s)}=\frac{\pm 1}{r(s)}\Rightarrow B(s)=\exp\left(\pm\int\frac{ds}{r(s)}\right).
$$
Also form (2) we get
$$
A(s)=\pm\int\sqrt{1-(B'(s))^2}ds=\pm\int\sqrt{1-\frac{B(s)^2}{r(s)^2}}ds=
$$
$$
=\pm\int\sqrt{1-\frac{\exp\left(\pm2\int r^{-1}(s)ds\right)}{r(s)^2}}ds
$$
Hence for a given $r(s)$, the curve can be parametrized as
$$
\overline{x}(s)=\left\{\pm\int\sqrt{1-\frac{1}{r(s)^2}\exp\left(\pm2\int\frac{ds}{r(s)}\right)}ds,\exp\left(\pm\int\frac{ds}{r(s)}\right)\right\},\tag 3
$$
for $s\in[a,b]$ and $s$ being its canonical parameter.