Equation of a Circle

Theorem

The equation of a circle with radius $R$ and center $(a,b)$ is

• $(x - a)^2 + (y - b)^2 = R^2$ in Cartesian coordinates
• $x = a + R\cos t, y = b + R\sin t$ as a parametric equation

In polar coordinates, it does not make sense to refer to a point by $x$ and $y$ coordinates. Instead, the center of a circle is commonly denoted $(r_0,\varphi)$, where $r_0$ is the distance from the origin and $\varphi$ is the angle from the polar axis in the counterclockwise direction. The equation for a circle with radius $R$ of this type is

• $r^2 - 2 r r_0 \cos(\theta - \varphi) + (r_0)^2 = R^2$

(note that $r$ is a function of $\theta$)

Corollary

If $(r_0,\varphi) = (0,0)$ (i.e. the circle is centered at the origin) this reduces to $r = R$

Proof

Cartesian

Let the point $(x,y)$ satisfy the equation $(x - a)^2 + (y - b)^2 = R^2$.

The distance between this point and the center of the circle is $\sqrt{(x - a)^2 + (y - b)^2}$ by the Distance Formula.

But from the equation, this quantity equals $R$, so the distance between points satisfying the eqn and the center is constant and equal to the radius.

$\blacksquare$

Parametric

Let the point $(x,y)$ satisfy the eqns $x = a + R\cos t, y = b + R\sin t$.

The distance between this point and the center of the circle is:

$\sqrt{[(a + R\cos t) - a]^2 + [(b + R\sin t) - b]^2}$

by the Distance Formula.

This simplifies to:

$\sqrt{R^2\cos^2 t + R^2\sin^2 t} = R\sqrt{\cos^2 t + \sin^2 t}$

Then by the Pythagorean trigonometric identity, this distance equals $R$, so the distance between points satisfying the eqn and the center is constant and equal to the radius.

$\blacksquare$

Polar

Let the point $(r,\theta)_\text{Polar}$ satisfy the eqn $r^2 - 2 r r_0 \cos(\theta - \varphi) + (r_0)^2 = R^2$.

The first thing we have to do is rewrite the points $(r,\theta)$ and $r_0,\varphi$ in Cartesian coordinates:

• $(r,\theta)_\text{Polar} = (r\cos\theta,r\sin\theta)_\text{Cartesian}$ and
• $(r_0,\varphi)_\text{Polar} = (r_0\cos\varphi,r_0\sin\varphi)_\text{Cartesian}$.

Thus the distance between the point $(r,\theta)_\text{Polar}$ the center of the circle is:

$\sqrt{(r\cos\theta - r_0\cos\varphi)^2 + (r\sin\theta - r_0\sin\theta)^2}$

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt{r^2\cos^2\theta + (r_0)^2\cos^2\varphi - 2r r_0\cos\theta\cos\varphi + r^2\sin^2\theta + (r_0)^2\sin^2\varphi - 2r r_0\sin\theta\sin\varphi}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt{r^2(\cos^2\theta + \sin^2\theta) + (r_0)^2(\cos^2\varphi + \sin^2\varphi) - 2 r r_0( \cos\theta\cos\varphi + \sin\theta \sin\varphi)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(=\)

\(\displaystyle \sqrt{r^2 + (r_0)^2 - 2 r r_0\cos(\theta - \varphi)}\)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

the trigonometric angle subtraction formula and the Pythagorean trigonometric identity

But from the equation, this quantity equals $R$, so the distance between points satisfying the equation and the center is constant and equal to the radius.

$\blacksquare$

Proof of Corollary

If $(r_0,\varphi) = (0,0)$, then the equation reduces to $r^2 = R^2$, and the result follows.

$\blacksquare$

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $10.15$