I have some problems understanding circles.

$x^2+y^2 = 1$ is a circle. It defines equivalence class where all (x,y) points belonging to the circle are in the same equivalence class. $(\cos a, \sin a)$ breaks that equivalence class, and allows us to enumerate the (x,y) points in the circle by varying the $a$ parameter. $x=\cos a, y=\sin a$ links $x^2+y^2=1$ and $(\cos a, \sin a)$ and does substitution to get $\cos(a)^2+\sin(a)^2=1$.

Another representation for the circle is $f(x,y)=c$ implicit function. Choosing $f(x,y)=x^2+y^2$ and $c=1$ brings us to the same circle as before. But what if we wanted $f$ to be some other function, and wanted to break the equivalence class in same way as with the case of circle. Basically, given unknown function $f(x,y)$, how can we find the representation like $(\cos a, \sin a)$ that allows breaking the equivalence class represented by $f(x,y)=c$ implicit function?

The obvious use case for this is that it allows us to use implicit functions without enumerating the whole (x,y) space and testing every point in the space whether it belongs to the subset. This requirement for enumerating the whole space comes from inverse image: $(k\rightarrow m) \rightarrow (m \rightarrow 2) \rightarrow (k \rightarrow 2)$, which is implemented simply by function composition. Applying this to the circle case gives us way to assign $k=(x,y)$, $m=R$ to get $((x,y) \rightarrow R) \rightarrow (R \rightarrow 2) \rightarrow ((x,y) \rightarrow 2)$. The main problem that this solution has, is that it requires enumerating the $(x,y)$ space to discover all values of $2$ which maps to each $(x,y)$ point, when we really wanted to enumerate just one dimension $a$.

So is there any ideas what methods there are to discover things like $(\cos a, \sin a)$ for generic implicit functions like $f(x,y)$? Obvious difficult case is $f(x,y) = g_1(x,y) + g_2(x,y)$.

  • $\begingroup$ I don't understand what you are trying to say with "breaks the equivalence class." Representing $x$ and $y$ in terms of a parameter $a$ does not remove $(x,y)$ from the circle, so they are still in the same equivalence class. $\endgroup$ Jun 16, 2016 at 4:58
  • $\begingroup$ I mean, I can think of the equivalence class of a circle just as one area, where different points of the circle cannot be distinguished from each other. But breaking it, means that it is divided to (small) parts which can be separated by parameter $a$. $\endgroup$
    – tp1
    Jun 16, 2016 at 5:46

1 Answer 1


I found myself one solution to the problem. It uses substitution $y=f(x)$, and keeps f(x) as unknown function, which can be calculated from the equation. Steps go as follows:

$$ f(x,y)=c $$ $$ f(x,f(x))=c $$ For circle, $f(x,y) = x^2+y^2$, so $$ x^2 + f(x)^2 = c $$ $$ f(x)^2 = c - x^2 $$ $$ f(x) = \sqrt{(c-x^2)} \vee f(x)=-\sqrt{(c-x^2)} $$

So basically this reduces the number of dimensions in the problem.


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