Circles radius $R$ drawn on parabola
$$ (x-2 at)^2+(y-at^2)^2=R^2\tag1$$
Differentiating with respect to $t$ and simplifying we get
$$ -4ax+ 8a^2t-4ya t +4a^2t^3=0; \rightarrow t^3+t(2-y/a)-x/a =0 \tag2$$
How can we now eliminate $t$ between third and and fourth order equations 1) and 2)?
Goebner basis could be applied I guess but how is it actually done? Two solutions are expected and how do they come about in the elimination?
Thanks for your help in advance.
The offset curve at distance $d$ can be found as the locus of the tip of a normal vector of length $d$.
$$x=2at,y=at^2\to\vec n=\pm\frac{2ad}{\sqrt{1+t^2}}(t,-1),$$ giving the parametric equations
$$x=2at\pm\frac{2ad}{\sqrt{1+t^2}}t,\\y=at^2\mp\frac{2ad}{\sqrt{1+t^2}}.$$
Elimination of $t$ doesn't seem easy.
Have a look at the incredible "Brief atlas of offset curves".