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I can't find the definition of a function $f(x); x \in [-1;1]$ where $(x|f(x))$ is a point on the unit-circle.

Can you please give me a hint?

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Background:

I want to render a filled circle (programatically) onto a screen using an iteration from $x-rad \rightarrow x+rad$ where $x$ is the x-position of the circle on the screen, and $rad$ is the radius.

Solution:

$f(x) = cos(sin^{-1}(x))$

seems to work pretty well. For the programmers under us, here is Python-code to demonstrate my approach:

import math  def drawCircle(xpos, ypos, radius):     for x in xrange(radius * 2):         x = x - radius          ytop = math.cos(math.asin( x / radius ))         ybot = -ytop          x = x + xpos         drawLine( x, ypos+ytop, x, ypos+ybot) 

Result:

image of an orange circle rendered using python

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    @AndréNicolas I didn't say, I wanted the function to replicate the whole unit-circle, the resulting point just needs to lie on it. Maybe my *edit* makes it more clear. Thanks to everyone who offered his help.2012-01-23

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For $-1\le x\le 1$, $\cos(\sin^{-1}x)=\sqrt{1-x^2}.$

A circle with radius $r$, centered at the origin can be described by $x^2+y^2=r^2.$ Solving for $y$ in terms of $x$ gives $y=\pm\sqrt{r^2-x^2}.$

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    Definately a nicer than using $cos$ and $sin$. I sure wouldn't have come to this equation. Thank you!2012-01-23