Equation of 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 (note that $$r$$ is a function of $$\theta$$)
 * $$r^2 - 2 r r_0 \cos(\theta - \varphi) + (r_0)^2 = R^2$$

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

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 equation and the center is constant and equal to the radius.

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

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

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

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

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

$$ $$

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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.

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