Given a probability density function $f(x)$ with support $[0,2\pi]$, I'm interested in constructing a "roulette" in which the "winning angles" follow that distribution. The shape of the roulette would be a curve $\Gamma$ described by $(cos(\theta)r(\theta), sin(\theta)r(\theta))$, and the length of $\Gamma$ between the angles $\theta_1$ and $\theta_2$ must be equal to the probability of the interval $[\theta_1, \theta_2]$.
$$\int_{\theta_1}^{\theta_2}{\sqrt{r(\theta)^2+r'(\theta)^2}d\theta} = \int_{\theta_1}^{\theta_2}{f(x)dx}$$
Then, one way to proceed is to, using the given $f$, find the solution to the non-linear differential equation.
$$r(x)^2+r'(x)^2 = f(x)^2$$
or
$$r(x)^2(1+(\frac{r'(x)}{r(x)})^2) = f(x)^2$$
From the latter one can get that an exponential $r$ is a solution when $f$ is exponential (because then the logarithmic derivative of $r$ is constant), but this is the only case I've been able to solve: using Fourier and Laplace transforms just gave equally hard (to me) questions about whether a function could be written as the integrand of a convolution of a function with itself.
Is there a general solution to the differential equation? Is there at least a solution when $f(x)=\frac{1}{2\Phi(\pi)-1}\frac{1}{\sqrt{2\pi}}e^{-\frac{(x-\pi)^2}{2}}$? What about an algorithm for generating approximations? For this I've tried replacing $f$ with a piecewise exponential function, but the resulting recurrence relation to make sure $r$ was continuous was beyond me.
Alternatively, is there another way of building the curve other than solving the equation?