Implicit representation of the involute of a circle Is there an implicit equation for the involute of a circle?
For example for a circle can be defined as $ x^2 + y^2 - r^2 = 0 $, but the only equations for the involute of a circle I can find are in parametric representation.
 A: One implicit representation is $f(x,y;a)=0$ where
$$
\textstyle f(x,y;a)=x\cos\left(\frac{1}{a}\sqrt{x^2+y^2-a^2}\right) 
+ y\sin\left(\frac{1}{a}\sqrt{x^2+y^2-a^2}\right) -a,
$$
and where $a$ is the radius of the circle that you're taking the involution of. In other words, for the involute of a unit circle we have
$$
\textstyle f(x,y) = x\cos(\sqrt{x^2+y^2-1})
+ y\sin(\sqrt{x^2+y^2-1}) -1,
$$
and for the involute of a general circle we have $f(x,y;a)=f(x/a,y/a)$.
Here is $f(x,y)$ plotted using Wolfram Alpha.
Source
Price, B (1865). A Treatise on Infinitesimal Calculus, Volume II: Integral Calculus, Calculus of Variations, and Differential Equations. 2nd ed. 
This is the sort of thing that was the bread and butter of geometry courses about 150 to 200 years ago, but is now out of fashion, so it's not a surprise that the reference is so old. Thank heavens for Google books!
The thing you really want: Link to relevant page in Google books.
The relevant equation is about halfway down page 227, in Example 1 of Article 172. Note that the author uses $\theta$ for the parameter, whereas the Wikipedia article uses $t$.
Derivation
The derivation is very short and easy (but not obvious if you haven't seen it!), so I'll copy it here. The parametric equations are 
$$x = a(\cos t + t\sin t),\quad y = a(\sin t - t\cos t).$$
As lab bhattacharjee's comment (and the Wikipedia article) says, you can add the square of these two together to get
$$x^2 + y^2 = a^2(1+t^2).$$
The trick from Price's book is to multiply the first equation by $\cos t$ and the second by $\sin t$ and add, giving
$$x \cos t + y \sin t = a.$$
Cancelling $t$ using these two equations gives you the answer.
A: Do you mean an equation like $p(x,y) = 0$, where $p$ is a polynomial in $x$ and $y$? I don't think so. Your curve should intersect any line through the origin in an infinite number of points, and this would contradict Bezout's theorem.
This doesn't rule out some sort of non-algebraic implicit equation, but I'm not sure what you would deduce from such a thing.
