# equation for the region inside a circle

What equation or group of equations fill the entire or part of a region inside a circle without using inequalities?

Update

I don't know if this problem is already solved, I'm trying to find the "length" of the region inside the circle. The function sould be continuos and with integrable length of arc.

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From your figure, it looks like you're looking for a parametric curve in polar coordinates, $r = f(\theta)$, whose range for $\theta\in[-\infty,\infty]$ is the entire circle/annulus, is that right? – Rahul Jan 1 '13 at 9:14
@JasperLoy I've updated the question for clarification – rraallvv Jan 1 '13 at 15:46
@RahulNarain I was thinking about a parametric function with correspondence one to one (binjective function), but my guess is that it can't be bijective. – rraallvv Jan 1 '13 at 15:47

Essentially, your question boils down to asking for equations whose solution set is bounded (since you can just divide by the sup, and multiply by the radius of your circle). There isn't a good answer, unless you can further restrict the functions that you're looking at - must they be continuous functions? polynomials?

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I was thinking about a parametric function with correspondence one to one (binjective function), but my guess is that it can't be bijective. – rraallvv Jan 1 '13 at 6:53
@rraallvv Is the parametric function going to be continuous (do you know what this word means?) How about $f: [ 0 , 1 ] \rightarrow \mathbb{R}^2$ given by $f (t) = \begin{cases} t & t \in \mathbb{Q} \\ -t & t \not \in \mathbb{Q}\\ \end{cases}$. – Calvin Lin Jan 1 '13 at 6:57
I don't know if this problem is already solved, I'm trying to find the "length" of the region inside the circle. The function sould be continuos and with a length of arc integrable. – rraallvv Jan 1 '13 at 7:24
There is such a thing as a space-filling curve, so you can get continuous solutions. But you won't find one where the arc length is finite, because a curve of finite arc length has Hausdorff dimension $1$, while the region inside a circle or an annulus has Hausdorff dimension $2$. – Robert Israel Jan 1 '13 at 9:28