Proving the curvature formula for an arbitrary planar curve using perpendicular bisectors

You have $Q = C(t)$ and $R = C(t + h) $

The center of the circle passing through these two points satisfies

$(S - Q) \cdot (S - Q) = (S - R) \cdot (S - R) $

so that

$ S \cdot S - 2 S \cdot Q + Q \cdot Q = S \cdot S - 2 S \cdot R + R \cdot R $

from which

$ 2 S \cdot (R - Q ) = R \cdot R - Q \cdot Q \hspace{15pt}(1) $

Using Taylor series approximation of $R$, we get

$R = C(t + dt) = C(t) + h C'(t) + \frac{1}{2} h^2 C''(t) $

Plugging this into $(1)$ and retaining only differentials of first and second order, we get

$ 2 S \cdot ( h C' + \frac{1}{2} h^2 C'' ) = 2 h C \cdot C' + h^2 (C \cdot C'' + C' \cdot C' ) $

From this, it follows that

$ S \cdot C' = C \cdot C' \hspace{15pt}(2)$

and

$ S \cdot C'' = C \cdot C'' + C' \cdot C'\hspace{15pt}(3) $

Equation $(2)$ can be written as

$ (S - C) \cdot C' = 0 \hspace{15pt}(4) $

So that $(S - C)$ is perpendicular to $C'$. It follows that

$ S = C + \alpha R(90^\circ) C' \hspace{15pt} (5) $

Explicitly writing the components of $C= (x, y)$, $C' = (x', y')$, and $C'' = ( x'', y'' ) $, then $(5)$ becomes

$ S = (x - \alpha y' , y + \alpha x' ) $

Plugging this into $(3)$

$ x'' (x - \alpha y') + y'' (y + \alpha x') = x x'' + y y'' + x'^2 + y'^2$

Cancelling equal terms on both sides of the equation and solving for $\alpha$

$\alpha = \dfrac{ x'^2 + y'^2 }{ x' y'' - y' x'' } $

Now the radius of the circle is $\| S - Q \| = \alpha \| C' \| = \alpha \sqrt{x'^2 + y'^2 } $

Hence

$ r = \dfrac{ (x'^2 + y'^2)^{\frac{3}{2}} } { x' y'' - y' x''} $

and the curvature is

$ \kappa = \dfrac{1}{r} =\dfrac { x' y'' - y' x''}{ (x'^2 + y'^2)^{\frac{3}{2}} } $

Using perpendicular bisectors we can define the osculating circle whose center $C$ is the center of curvature and whose radius is $\rho$.

The circle lying in the osculating plane at the curve at the point $P$, having radius $\rho$ is called the "osculating circle". We can show that: "the osculating circle is the limit figure of a circle passing through three points of the given curve, when they tend to coincide in only one. Let the center of this last circle be $C$, its radius $r$ and $P$, $P_{1}$, $P_{2}$ the three points that tend, in the passage to the limit, to coincide with $P$. Setting

$F(t)=(C-M)^{2}-r^{2}$,
which represents the equation of a circle of center $C$ and radius $r$ , passing through the generic point $M$.

Then the function $F(t)$ must be null when instead of $M$ we place $P$, $P_{1}$, $P_{2}$, since it is null when $M$ is a point of the circle of center $C$ and radius $r$.

By Rolle's theorem $F(t)$ is null at a point $P_{1}*$ between $P$ and $P_{1}$ and at a point $P_{2}*$ between $P_{1}$ and $P_{2}$;

and again for the same theorem $F''(t)$ is null at a point $P*$ between $P_{1}*$ and $P_{2}*$.

When $P_{1}$ and $P_{2}$ tend to $P$, $P_{1}*$, $P_{2}*$ and $P*$ also tend to $P$, then the center $C$ is defined by the equations:

$F(t)=0$, $F’(t)=0$, $F’’(t)$ ;

i.e.

$(C-P)^{2}=r^{2}$,

$(C-P)\dot{P}=0$,

$(C-P)\ddot{P}=\dot{P}^{2}$.

The first equation is the equation of the center circle $C$ and radius $r$, the second says that $C$ lies in the plane normal to the curve in $P$; therefore $C$ is on the main normal, because, having to be coplanar with $P$, $P_{1}$, $P_{2}$ at the limit, it must be on the osculating plane.

The third equation taking as a parameter the arc $s$, becomes:

$(C-P) \frac{1}{\rho} \vec{n}=1$,

i.e. remembering that $r$ is the module of $C-P$ e che $C-P$ ed $\vec{n}$ are parallel:

$r=\rho$.

At this point referring to Prove a relation for radius of curvature, using parametric coordinates

and in particular the formula:

$\frac{1}{\rho^{2}}= \frac{\ddot{x}^2+\ddot{y}^2-\ddot{s}^2}{\dot{s}^4}$

Placing in place of

$\ddot{s}=\frac{\dot{x}\ddot{x}+\dot{y}\ddot{y}}{\dot{s}}$,

and in place of

$\dot{s}^2=\dot{x}^2+\dot{y}^2$,

we get:

$\frac{1}{\rho}=\frac{\dot{x}\ddot{y}-\dot{y}\ddot{x}}{\Big(\dot{x}^2+\dot{y}^2\Big)^{\frac{3}{2}}}$.