Circle of curvature

Center of curvature^14.1. If a circle be drawn through three points $P_0$, $P_1$, $P_2$ on a plane curve, and if $P_1$ and $P_2$ be made to approach $P_0$ along the curve as a limiting position, then the circle will in general approach in magnitude and position a limiting circle called the circle of curvature of the curve at the point $P_0$. The center of this circle is called the center of curvature.

Figure 14.1: Geometric visualization of the circle of curvature.

Let the equation of the curve be

$$y = f(x);$$ (14.1)

and let $x_0$, $x_1$, $x_2$ be the abscissas of the points $P_0$, $P_1$, $P_2$ respectively, $(\alpha',\alpha')$ the coordinates of the center, and $R'$ the radius of the circle passing through the three points. Then the equation of the circle is

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

and since the coordinates of the points $P_0$, $P_1$, $P_2$ must satisfy this equation, we have

$$\begin{cases}(x_0 - \alpha')^2 + (y_0 -\beta')^2 - R'^2 = 0, ... ...\ (x_2 - \alpha')^2 + (y_2 - \beta')^2 -(R')^2 = 0. \end{cases}$$ (14.2)

Now consider the function of $x$ defined by

$$F(x) = (x-\alpha')^2 + (y-\beta')^2 - (R')^2,$$

in which $y$ has been replaced by $f(x)$ from (14.1).

Then from equations (14.2) we get

$$F(x_0) = 0,\ F(x_1) = 0,\ F(x_2) = 0.$$

Hence, by Rolle's Theorem (§13.1), $F'(x)$ must vanish for at least two values of $x$, one lying between $x_0$ and $x_1$, say $x'$, and the other lying between $x_1$ and $x_2$ say $x''$; that is,

$$F'(x') = 0,F'(x'') = 0.$$

Again, for the same reason, $F''(x)$ must vanish for some value of $x$ between $x'$ and $x''$, say $x_3$; hence

$$F''(x_3) = 0.$$

Therefore the elements $\alpha'$, $\beta'$, $R'$ of the circle passing through the points $P_0$, $P_1$, $P_2$ must satisfy the three equations

$$F(x_0) = 0,\ F'(x') = 0,\ F''(x_3) = 0.$$

Now let the points $P_1$ and $P_2$ approach $P_0$ as a limiting position; then $x_1$, $x_2$, $x'$, $x''$, $x_3$ will all approach $x_0$ as a limit, and the elements $\alpha$, $\beta$, $R$ of the osculating circle are therefore determined by the three equations

$$F(x_0) = 0,\ F'(x_0) = 0,\ F''(x_0) = 0;$$

or, dropping the subscripts, which is the same thing,

$$(x - \alpha)^2 + (y -\beta)^2 = R^2$$ (14.3)

$$(x - \alpha) + (y -\beta)\frac{dy}{dx} = 0,$$ (14.4)

differentiating (14.3).

$$1 + \left( \frac{dy}{dx} \right)^2 + (y - \beta) \frac{d^2 y}{dx^2} = 0,$$ (14.5)

differentiating (14.4). Solving (14.4) and (14.5) for $x-\alpha$ and $y-\beta$, we get $\left( \frac{d^2 y}{dx^2} \ne 0 \right)$,

$$\begin{cases}x - \alpha = \frac{\frac{dy}{dx} \left[ 1 + \lef... ...left( \frac{dy}{dx} \right)^2}{\frac{d^2 y}{dx^2}}; \end{cases}$$ (14.6)

hence the coordinates of the center of curvature are

$$\alpha = x - \frac{\frac{dy}{dx} \left[1 + \left( \frac{dy}{dx} \... ...dy}{dx} \right)^2}{\frac{d^2 y}{dx^2}}. \left( \frac{d^2 y}{dx^2} \ne 0 \right)$$ (14.7)

Substituting the values of $x-\alpha$ and $y-\beta$ from (14.6) in (14.3), and solving for $R$, we get

$$R = \pm \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{3}{2}} }{ \frac{d^2 y}{dx^2} },$$

which is identical with (12.5), [§12.5]. Hence

Theorem 14.1.1   The radius of the circle of curvature equals the radius of curvature.