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.
 A: Actually the answer comes out more "fluidly", when you 
consider the "mechanical" construction of the involute, as the trace of the end point (call it $P=(x,y)$) of a tether which is evolving from a circle of radius $r$.
The angle $\theta$ in the parametric equation you shown is in fact the trigonometric angle that the tether has unwound: call $T=(r\; cos\theta, r\; sin\theta)$ the corresponding point on the circle. Then:


*

*$\left| {TP} \right| = r\;\theta $

*$TP$ is tangent to the circle in $T$

*by construction, $T$ is the center of curvature of the involute at $P$

*the line through $T$ and $P$ is thus normal to the involute (in $P$)  


So, given a point $A=(a,b)=(\rho\; cos\alpha,\rho\; sin\alpha)$ external to the circle (re. to this figure)


*

*draw the tangent from $A$ to the circle, on the side where $A$ will remain on the same side as $P$, with respect to $T$.

*then you get $\theta  = \alpha  + \arccos \left( {r/\rho } \right)$ and $\left| {TA} \right| = \sqrt {\rho ^{\,2}  - r^{\,2} } $

*the tangent line will cross the involute orthogonally at all the various points, whose distance from $T$ is 
$r\,\left( {\theta  + 2\,k\,\pi } \right)\quad \left| \begin{gathered}
  \;0 \leqslant \theta  < 2\pi  \hfill \\
  \;1 \leqslant k \hfill \\ 
\end{gathered}  \right.$  


Finally, $\left| {TA} \right| - r\;\theta $ will give you the outward distance from the first turn of the involute. If positive, when reduced $\bmod \left( {2\,\pi \,r} \right)$ will give the distance from nearest inner turn.
Refer also to this post.  
Addendum 


*

*which tangent from $A$ to the circle shall be choosen
As the tether unwinds from the circle counter-clockwise, the straighten portion will remain "to the right" of the tangent point, for an observer standing vertically at the origin. The point $A$ shall appear on the same side.
Mathematically speaking the tangent to choose shall be the one for which
$0 \leqslant \left| {\mathop {TA}\limits^ \to  \; \times \;\mathop {OT}\limits^ \to  } \right| $.
That translates in just choosing the positive value for $\arccos \left( {r/\rho } \right)$.  

*Case in which the point $A$ is internal to the circle
When the point $A=(a,b)=(\rho\; cos\alpha,\rho\; sin\alpha)$ is inside
the circle, i.e. $\rho  < r$, the previous formulas do not hold.
The normal lines to the involute are all tangent to the circle (that's why this curve is used in the gears tooth profile),
so there is no normal through $A$ to the curve, hence no local point of minimum (or max) distance.
Since the distance $\left| {AP} \right|$ is a continuous function in $\theta$, then the minimum shall occur at the extremal points.
To this purpose we can limit and consider the portion of the curve for  $0 \leqslant \theta  < 2\,\pi $ (first turn). 
The distance between $A$ and the extreme point at $\theta=0$  ($P(0)=(r,0)$) will be at most $2r$ .
The distance between $A$ and the extreme point at $\theta=2\pi$  ($P(2\pi)=(r,2r\pi)$) will be at least $\left( {\sqrt {1 + 4\pi ^2 }  - 1} \right)r .$
Thus the minimum distance will be that between $A$ and the extreme point $U=(r,0)$.
To demonstrate that algebrically, we shall show that
$$
\left| {UA} \right| \leqslant \left| {AP} \right|\quad \left| \begin{gathered}
  \;\rho  < r, \hfill \\
  \;0 \leqslant \alpha ,\theta  < 2\,\pi  \hfill \\ 
\end{gathered}  \right.
$$
i.e.
$$
\left( {\rho /r\cos \alpha  - 1} \right)^2  + \left( {\rho /r\sin \alpha } \right)^2  \leqslant \left( {\cos \theta  + \theta \sin \theta  - \rho /r\cos \alpha } \right)^2  + \left( {\sin \theta  - \theta \cos \theta  - \rho /r\sin \alpha } \right)^2 
$$
putting:
$$
\left\{ \begin{gathered}
  0 \leqslant \theta  < 2\,\pi  \hfill \\
  0 \leqslant \beta  = \alpha  - \theta  < 2\,\pi  \hfill \\
  0 \leqslant \nu  = \rho /r \leqslant 1 \hfill \\ 
\end{gathered}  \right.
$$
it comes to
$$
0 \leqslant \theta ^{\,2}  + 2\,\nu \left( {\sin \beta } \right)\,\theta  + 2\,\nu \,\cos \left( {\beta  + \,\theta } \right) - 2\,\nu \cos \beta 
$$
which is not an easy task (could maybe the subject of a dedicated post).

*Local vs. absolute minimum
The considerations made above, relying on individuating the center of curvature and normal line to the curve, leed to find the points on the curve for which $\left| {AP} \right| (\theta)$ has a local minimum. When $A$ is internal to the first turn of the involute, the starting value $\left| {AP} \right| (0)= \left| {UA} \right|$ might well be below that, and shall be checked for, if the absolute minimum distance is to be found. 

A: 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:

$$TQ^2 = a^2+b^2-r^2 ;\, QI = r \theta - TQ = r \theta - \sqrt{  a^2+b^2-r^2 }. $$
A: The combination of a Newton-Raphson method together with the suggestion by @Alexey Burdin worked quite efficiently. 
