# How do you find the parametric equations of a circle?

##### 1 Answer
Aug 23, 2014

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

$y = r \sin t$
$x = r \cos t$

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

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

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

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

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

$y = r \sin \arccos \left(\frac{x}{r}\right)$

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

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

$y = r \cdot \frac{\sqrt{{r}^{2} - {x}^{2}}}{r}$

This simplifies to

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

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

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

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