Evolute of Circle is its Center

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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.

$\blacksquare$


Sources

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