How do you find the parametric equations of a circle?

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How do you find the parametric equations of a circle?

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Answer 1 · 2014-08-23T21:39:16

We'll start with the parametric equations for a circle:

$y = r sin t$ $y = rsin t$
$x = r cos t$ $x = rcos t$

where $t$ $t$ is the parameter and $r$ $r$ is the radius.

If you know that the implicit equation for a circle in Cartesian coordinates is $x^{2} + y^{2} = r^{2}$ $x^2 + y^2 = r^2$ then with a little substitution you can prove that the parametric equations above are exactly the same thing.

We will take the equation for $x$ $x$ , and solve for $t$ $t$ in terms of $x$ $x$ :

$\frac{x}{r} = cos t$ $x/r = cos t$
$t = arccos (\frac{x}{r})$ $t = arccos (x/r)$

Now substitute into the equation for $y$ $y$ . This eliminates the parameter $t$ $t$ and gives us an equation with only $x$ $x$ and $y$ $y$ .

$y = r sin arccos (\frac{x}{r})$ $y = rsin arccos(x/r)$

$sin arccos (\frac{x}{r})$ $sin arccos(x/r)$ is equal to $\frac{\sqrt{r^{2} - x^{2}}}{r}$ $sqrt(r^2 - x^2)/r$ . This is apparent if one sketches a right triangle, letting $θ = arccos (\frac{x}{r})$ $theta = arccos(x/r)$ :

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Thus, $sin θ = \frac{\sqrt{r^{2} - x^{2}}}{r}$ $sin theta = sqrt(r^2 - x^2)/r$ . So now we have

$y = r \cdot \frac{\sqrt{r^{2} - x^{2}}}{r}$ $y = r*sqrt(r^2 - x^2)/r$

This simplifies to

$y = \sqrt{r^{2} - x^{2}}$ $y = sqrt(r^2 - x^2)$

If we square this entire deal and solve for $r$ $r$ , we get:

$r^{2} = x^{2} + y^{2}$ $r^2 = x^2 + y^2$

which is precisely the equation for a circle in Cartesian coordinates.

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How do you find the parametric equations of a circle?

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