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 $r=1$:

enter image description here

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!

  • 1
    $\begingroup$ If I'm not mistaken, the circle is chosen so that its second derivative is equal to that of the graph. That might just be for this special case where the circle is being fitted at a stationary point. The idea is to find the circle which best approximates the graph in a small region. We can get any circle to have the same first derivative-- we tune its radius to get its second derivative to match the graph. $\endgroup$ – Myridium Jan 9 '18 at 12:31

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


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


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


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 $y=\frac{1}{2}x^2$.

  • $\begingroup$ I like this, but would like a bit more. Without knowing the circle beforehand, how would you proceed? $\endgroup$ – Forever Mozart Jan 9 '18 at 0:13
  • $\begingroup$ You could try a circle of radius $r$: $x^2 + (y-r)^2 = r^2$, and see what $r$ has to be... $\endgroup$ – Robert Israel Jan 9 '18 at 0:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.