Distance from a point to the involute of a circle I know that the involute of circle of radius $r$ centered at $(0,0)$ is given by the following parametric form:
$$\begin{cases}
x(\theta) = r \big(\cos(\theta) + \theta\ \sin(\theta) \big),\\
y(\theta) = r\big(\sin(\theta) - \theta\ \cos(\theta) \big),
\end{cases}$$
with $\theta\in\mathbb{R}$. Given a point $(a,b)\in\mathbb{R}^2$, I would like to compute its distance to the involute (or the closest point on the involute).
Given the parametric form I can compute the normal direction to the involute at a point and, thus, I need to compute a $t\in\mathbb{R}$ such that 
$$(a,b) = \big(x(\theta),y(\theta)\big) \ + \ t\ \big(-y'(\theta),x'(\theta)\big).$$
I guess that now the easiest way consists on computing $\theta$ first, which gives the closest point:
$$\frac{x(\theta)-a}{y'(\theta)}-\frac{b-y(\theta)}{x'(\theta)}=0.$$
It is obvious that one needs iterative solvers to approximate the solution (I have already tried Newton-Raphson).
The main problem is that the function to be minimized is highly oscillating and it can be difficult to give a good approximation for the initialization of the algorithm. In fact, it seems that this should also depend on the radius $r$.
Any help or suggestion is really welcome: from a geometrical point of view (a way to simplify the computations, an already computed formula...), from a numerical point of view (other algorithms) or from the implementation point of view.

We can use property of involute that the tangent to base circle is normal to involute.
Let given point be Q. Draw tangent to circle radius $a$ on concave side. Let the tangent  point be  $T$. Extend QT to cut given involute at $I$ ( automatically cuts normal to involute if convex).
Find required segment of tangent length as $IQ$
The method is suitable even when Q lies on convex side, as the point Q is outside the involute.
EDIT1:

$$TQ^2 = a^2+b^2-r^2 ;\, QI = r \theta - TQ = r \theta - \sqrt{  a^2+b^2-r^2 }. $$
