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 = rsin t#
#x = rcos 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#:

#x/r = cos t#
#t = arccos (x/r)#

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

#y = rsin arccos(x/r)#

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

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Thus, #sin theta = sqrt(r^2 - x^2)/r#. So now we have

#y = r*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.