Parametric Equations of a Circle

The Cartesian equation $(x-h)^2+(y-k)^2=r^2$ describes a circle implicitly — it tells you which $(x,y)$ pairs lie on the curve, but not how to sweep through them one at a time. The parametric form fixes that by describing every point on the circle using a single angle parameter $\theta$:

Geometrically, $\theta$ is the angle that the radius to the point $(x,y)$ makes with the positive-$x$-direction line through the centre. As $\theta$ runs once around from $0$ to $2\pi$, the point $(x,y)$ traces the entire circle exactly once. You can check the form is consistent with the Cartesian equation by substituting back: $(x-h)^2+(y-k)^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2(\cos^2\theta+\sin^2\theta)=r^2$, using the Pythagorean identity. For a circle centred at the origin this simplifies to $x=r\cos\theta$, $y=r\sin\theta$.

The real payoff of the parametric form is that it turns a two-variable geometry problem into a one-variable trigonometry problem — very useful for finding the greatest or least value of some linear expression on a circle, or for describing motion around a circular path.

Worked Example. Parametrise the circle $x^2+y^2-6x+8y+9=0$, and use the parametrisation to find the maximum value of $x+y$ on this circle.

First put the equation in general form to read off the centre and radius: $2g=-6\Rightarrow g=-3$, $2f=8\Rightarrow f=4$, $c=9$, so the centre is $(-g,-f)=(3,-4)$ and the radius is $\sqrt{9+16-9}=\sqrt{16}=4$. The parametric equations are therefore $x=3+4\cos\theta,\ y=-4+4\sin\theta$. …