Approximating graph with circles... elementary method?

Question

Let $f:\mathbb R \to \mathbb R$ a twice continuously differentiable function.

According to wikipedia https://en.wikipedia.org/wiki/Curvature, the radius of the approximating circle that touches the curve $y=f(x)$ at the point $\langle a,f(a)\rangle$ is given by:

$${r={\frac{\left(1+[f'(a)]^{2}\right)^{\frac {3}{2}}}{f''(a)}}.}$$

If $f(x)=\frac{1}{2}x^2$ then: $${r={\left(1+a^2\right)^{\frac {3}{2}}}.}$$

So if $a=0$ then we get $r=1$:

Question. Is there an elementary way (using only simple geometry and limits) to prove this particular radius (with $f(x)=\frac{1}{2}x^2$ and $a=0$) is equal to $1$?

EDIT: To be clear, I'd like a solution that takes limits of circles through points near $\langle 0,0\rangle$. Just like with tangent lines. Surely this can be done!

Answer 1

The circle is described by the implicit equation $x^2+(y-1)^2=1$. The parabola is described by $y=\frac{1}{2}x^2$. When $y<<1$, we have that $(y-1)^2\approx 1-2y$, so the circle can be approximated as $x^2+1-2y=1$, or equivalently $y=\frac{1}{2}x^2$.

Answer 2

Consider the circle through $(0,0),(-t,t^2/2)$ and $(t,t^2/2)$, which has its center at $(0,r)$.

Hence

$$((\pm t)^2-0)+\left(\dfrac{t^2}2-r\right)^2=r^2,$$

$$\left(\frac{t^2}4-r+1\right)\,t^2=0$$

and $r$ tends to $1$ when $t$ tends to $0$.