Curvature Using Circles Given the equation $(x - h)^2 + (y - k)^2 = r^2$ representing the family of all circles of radius r at the point $(h,k)$ if we try to form the differential equation representing this family we find an equation of the form
$$\kappa = \frac{1}{r} = \frac{y''}{\sqrt{(1 + y'^2)^3}}$$
which is surprisingly the equation for the curvature of a plane curve (ignoring absolute values which arise in the derivation).
But this expression was derived interpreting $y$ as part of a circle, how in the world can one justify plugging in other functions & saying this represents the curvature of that curve at the point? I know you're trying to say that, locally, the curve moves as though it were moving along the arc of a circle of radius $r$, but there seems to be a jump in my mind as to how you get that interpretation out of what I've written.
 A: One definition of the curvature of a plane curve at a given point is $\tfrac{1}{\rho}$ where $\rho$ is the radius of the osculating circle to the curve at that point. 
Consider a smooth curve in the plane, say $C$, and a fixed point on that curve, say $p$. There are lots of circles tangent to $C$ at $p$. In fact, you find that all of these circle have one important thing in common: all of their centres lie on the normal line to $C$ at $p$. (If you draw the tangent line to $C$ at $p$ then the normal line is the line through $p$ that is perpendicular to this tangent line.) 
One of these tangent circles is different to all of the rest: it has higher order contact with $C$ at $p$. All but one of the circles are simply tangent to $C$. They meet $C$ like the the parabola $y=x^2$ meets the $x$-axis at the origin. As far as the first derivative is concerned, the circle and the curve are the same at $p$. 
A single circle, called the osculating circle, has higher order contact with $C$ at $p$. Ordinarily, it meets $C$ like $y=x^3$ meets the $x$-axis at the origin. As far as the first two derivatives are concerned, the circle and the curve are the same at $p$. 
Although, if $p$ is a vertex ($\kappa \neq 0$ and $\kappa' = 0$) then it meets $C$ like $y=x^4$ meets the $x$-axis at the origin. As far as the first three derivatives are concerned, the circle and the curve are the same at $p$.
(When I say "meets like", I mean up to diffeomorphism, i.e. up to smooth changes of coordinates.) 
The radius of this osculating circle, $\rho$, is called the radius of curvature of $C$ at $p$ and $\tfrac{1}{\rho}$ turns out to be $\kappa.$ Interestingly, if $p$ is an ordinary inflection then $\kappa = 0$ (and $\kappa' \neq 0$) and so $\rho = \infty$. The osculating circle is centred at infinity and the circle becomes a line, i.e. the curve meets it tangent line line $y=x^3$ meets the $x$-axis at the origin.
The beauty of working with the osculating circles and not the horrible formula for $\kappa$ is that they are totally independent of coordinates. There is something natural and uncontrived about the contact between circles and curves. Moreover, circles and lines are the orbits of Euclidean transformations. 
A: Try
$$ \kappa = \frac{1}{r} = \frac{ \dot{y} \ddot{x} - \ddot{y} \dot{x} }{ \left( \dot{x}^2 + \dot{y}^2 \right)^\frac{3}{2} }  $$
