What is the equation of a circle of radius say $\sqrt{40}$ moving on a parabolic path of equation say $f(x)=1/400 x^2$? ( Using Desmos) https://www.desmos.com/calculator/uug5sg29jt
I'd like to draw using Desmos a circular object moving on a parabolic path of equation , say : $f(x)=\frac{1}{400}x^2$.
Note : I'd like the circle not to have its center on the parabola itself, but at a distance of , say, $\sqrt {40}$ units.
I thought it would work if I located the center of the circle on a line perpendicular to the tangent of the parabola at a variable point $P=(a,f(a))$, $a$ being itself a varying number. .
I defined this perdendicular line as a function $g(x)$ and I used some trigonometry to obtain, from the slope of the perpendicular to the tangent,  the coordinates of the center of the desired moving circle.
What I obtain is a circle that actually slides on the parabola for $x\lt 0$ but that lies below the parabola for $x\gt 0$.
Could you please explain me what is going on at $x=0$ and after?

 A: Note that range of $tan^{-1}x$ is $(-\frac {\pi}{2}, \frac {\pi}{2})$, in which interval $\cos x$ is always positive. Hence, the $x$-coordinate of the centre of the circle, $c$, is always greater than $a$. Now, for $x>0$, slope of normal is negative, which means that if $c>a$, then $ g(c)<g(a)$. Since $g(a)$ lies on the parabola itself, hence the circle lies below the parabola. For $x<0$, since slope of normal is positive, this problem is not faced, as $g(c)>g(a)$ for that region.
To overcome this you should define $c$ separately taking positive and negative cases for $x$.
A: If you parametrize the parabola as $\displaystyle (t,\frac{t^2}{400})$ then,
$\displaystyle y' = \frac{t}{200}$ and so for the orthogonal line to the tangent, the slope is given by $y' = \displaystyle \tan\theta = - \frac{200}{t}$.
Now use $\sin\theta = \frac{|\tan \theta|}{\sqrt{1+\tan^2\theta}}, \cos\theta = \pm\frac{1}{\sqrt{1+\tan^2\theta}}$ to evaluate vertical and horizontal component of distance of $\sqrt{40}$.
So the center of the circle is,
$ \displaystyle \left(x_0 = t\ +\ \frac{\sqrt{40}t}{\sqrt{t^{2}+200^{2}}}, y_0 = \frac{t^{2}}{400}-\frac{200\sqrt{40}}{\sqrt{t^{2}+200^{2}}}\right)$
Equation of the circle is then $\displaystyle (x-x_0)^2 + (y-y_0^2) = 40$
Please note the center of the circle is outside the parabola. If I see your diagram, you want it inside the parabola for $t \lt 0$. If so, change the sign accordingly.
