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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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  • $\begingroup$ What does $(x|f(x))$ mean? Did you intend that to be $(x,f(x))$ a co-ordinate on the unit circle? $\endgroup$
    – user21436
    Jan 23, 2012 at 20:56
  • $\begingroup$ Draw the unit circle on a piece of paper, pick a point on it, consider its projections on the axes, and apply Pythagoras' rule. $\endgroup$
    – Emre
    Jan 23, 2012 at 20:57
  • $\begingroup$ Do you know of any functions that map to the unit circle? You knew enough to tag this "trigonometry", so it should come as no surprise if trig ratios come into play. $\endgroup$
    – The Chaz 2.0
    Jan 23, 2012 at 20:59
  • $\begingroup$ @Emre Oh man, simpler than I thought. Headache, you know? ;-) $f(x) = cos(sin(x))$ ? $\endgroup$
    – Niklas R
    Jan 23, 2012 at 21:02
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    $\begingroup$ Sorry, mistyped. I meant $f(x) = cos(sin^{-1}(x))$ which actually seems to work well. $\endgroup$
    – Niklas R
    Jan 23, 2012 at 21:36

1 Answer 1

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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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  • $\begingroup$ Definately a nicer than using $cos$ and $sin$. I sure wouldn't have come to this equation. Thank you! $\endgroup$
    – Niklas R
    Jan 23, 2012 at 22:28

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