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:

$k = \dfrac {y''} {\paren {1 + y'^2}^{3/2} }$

Here we have:

 $\ds 2 x + 2 y \frac {\d y} {\d x}$ $=$ $\ds 0$ differentiating with respect to $x$ $\ds \leadsto \ \$ $\ds \frac {\d y} {\d x}$ $=$ $\ds -\frac x y$ $\ds \leadsto \ \$ $\ds \frac {\d^2 y} {\d x^2}$ $=$ $\ds \frac {x \frac {\d y} {\d x} - y} {y^2}$ Quotient Rule for Derivatives $\ds$ $=$ $\ds \frac {y - x \paren {-\frac x y} } {y^2}$ substituting for $\dfrac {\d y} {\d x}$ $\ds$ $=$ $\ds \frac {y^2 + x^2} {y^3}$ simplifying

So:

 $\ds k$ $=$ $\ds \dfrac {\frac {y^2 + x^2} {y^3} } {\paren {1 + \paren {-\frac x y}^2}^{3/2} }$ $\ds$ $=$ $\ds \dfrac {y^2 + x^2} {y^3 \paren {1 + \frac {x^2} {y^2} }^{3/2} }$ $\ds$ $=$ $\ds \dfrac {y^2 + x^2} {y^3 \paren {\frac {y^2 + x^2} {y^2} }^{3/2} }$ $\ds$ $=$ $\ds \dfrac {y^2 + x^2} {\paren {y^2 + x^2}^{3/2} }$ $\ds$ $=$ $\ds \dfrac 1 {\paren {y^2 + x^2}^{1/2} }$ $\ds$ $=$ $\ds \dfrac 1 a$ 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.

$\blacksquare$