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.
then circles are given by $$(x-a)^2 + (y-a)^2 = 1, a \text{ arbitrary}. \tag 1 $$
differentiating $(1)$ with respect to $x,$ we have $$(x-a) +(y-a)y' = 0\tag 2 $$ solving $(2)$ for $a,$ we get $$a = \frac{x+yy'}{1+y'}, \quad x - a=\frac{(x-y)y'}{1+y'}, \quad y - a =\frac{y-x}{1+y'}\tag 3 $$
subbing $(3)$ in $(1)$ gives you $$(x-y)^2\left(y'^2+1\right) = (1+y')^2 $$
Best Answer
Take $(x-a)^2+(y-b)^2=r^2$ as the most general form of circles with center $(a,b)$ and radius $r$. Because of having three free parameters $a,b,r$ we must finally arrive at a 3-order differential equation by successive differentiation of the equation of the circles to eliminate all those constants so we start. $$(x-a)^2+(y-b)^2=r^2$$ $$2(x-a)+2y^{'}(y-b)=0$$ $$2+2(y^{'})^2+2y^{"}(y-b)=0$$ which by rearranging the terms leads to: $$b=\frac{1+yy^{"}+(y^{'})^2}{y^{"}}=\frac{1}{y^{"}}+y+\frac{(y^{'})^2}{y^{"}}$$ by third differentiation we finally arrive at: $$\frac{-y^{'''}}{(y^{"})^2}+3y^{'}-(\frac{y^{'}}{y^{"}})^2y^{'''}=0$$ $$\Large y^{'''}=\frac{3y^{'}}{1+(y^{'})^2}(y^{"})^2$$