# Evolute of Circle is its Center

## Theorem

The evolute of a circle is a single point: its center.

## Proof

By definition, the evolute of $C$ is the locus of the centers of curvature of each point on $C$

Without loss of generality, take the circle $C$ of radius $a$ whose center is positioned at the origin of a cartesian plane.

From Equation of Circle, $C$ has the equation:

- $x^2 + y^2 = a^2$

From the definition of curvature in cartesian form:

- $k = \dfrac {y''} {\left({1 + y'^2}\right)^{3/2} }$

Here we have:

\(\displaystyle 2 x + 2 y \frac {\mathrm d y} {\mathrm d x}\) | \(=\) | \(\displaystyle 0\) | differentiating with respect to $x$ | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \frac {\mathrm d y} {\mathrm d x}\) | \(=\) | \(\displaystyle -\frac x y\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \frac {\mathrm d^2 y} {\mathrm d x^2}\) | \(=\) | \(\displaystyle \frac {x \frac {\mathrm d y} {\mathrm d x} - y} {y^2}\) | Quotient Rule for Derivatives | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {y - x \left({-\frac x y}\right) } {y^2}\) | substituting for $\dfrac {\mathrm d y} {\mathrm d x}$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {y^2 + x^2} {y^3}\) | simplifying |

So:

\(\displaystyle k\) | \(=\) | \(\displaystyle \dfrac {\frac {y^2 + x^2} {y^3} } {\left({1 + \left({-\frac x y}\right)^2}\right)^{3/2} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {y^2 + x^2} {y^3 \left({1 + \frac {x^2} {y^2} }\right)^{3/2} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {y^2 + x^2} {y^3 \left({\frac {y^2 + x^2} {y^2} }\right)^{3/2} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {y^2 + x^2} {\left({y^2 + x^2}\right)^{3/2} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac 1 {\left({y^2 + x^2}\right)^{1/2} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac 1 a\) | as $x^2 + y^2 = a^2$ |

Thus the curvature of $C$ is constant.

The radius of curvature of $C$ is likewise constant:

- $\rho = a$

From Radius at Right Angle to Tangent, the normal to $C$ at all points on $C$ passes through the center of $C$.

We have that $a$ is the distance from $C$ to the center of $C$.

Thus it follows that the center of curvature of $C$ is the center of $C$ at all points.

Hence the result by definition of evolute.

$\blacksquare$

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {B}.23$: Evolutes and Involutes. The Evolute of a Cycloid