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:

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

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.

$\blacksquare$


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