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.