Let $S=0$ be the circle whose centre is at $\left(h,k\right)$and which cuts the parabola ${\mathrm{y}}^2=4\mathrm{ax}$  at four points $A,B,C,D$ so that triangle $ABC$ is equilateral. The fourth vertex can be 

Equation of circle  ${\left(\mathrm{x}-\mathrm{h}\right)}^2+{\left(\mathrm{y}-\mathrm{k}\right)}^2={\mathrm{r}}^2........\left(1\right)$
Substituting  $\left({\mathrm{at}}^2,2\mathrm{at}\right)\mathrm{in}......\left(1\right)$
${\left({\mathrm{at}}^2-\mathrm{h}\right)}^2+{\left(2\mathrm{at}-\mathrm{k}\right)}^2={\mathrm{r}}^2,$  gives the four values of t which represent the points A,B,C,D
 $\Rightarrow \sum {\mathrm{t}}_{\mathrm{i}}=0\sum {\mathrm{t}}_1{\mathrm{t}}_{\mathrm{j}}=4-\frac{2\mathrm{h}}{\mathrm{a}}........\left(2\right)$
 $\mathrm{a}\left({\mathrm{t}}_1^2+{\mathrm{t}}_2^2+{\mathrm{t}}_3^2\right)=3\mathrm{h},2\mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2,{\mathrm{t}}_3\right)=3\mathrm{k}$
 $\left(\because \mathrm{circumcentre}=\mathrm{centroid}\right)$
${\mathrm{t}}_4=\frac{-3\mathrm{k}}{2\mathrm{a}},\sum {\mathrm{t}}_1^2=-2\left(4-\frac{2\mathrm{h}}{\mathrm{a}}\right)\left(\because \mathrm{from}\left(2\right)\right)$   $\Rightarrow {\mathrm{at}}_4^2=\mathrm{h}-8\mathrm{a}$
$\left({\mathrm{at}}_4^2,2{\mathrm{at}}_4\right)=\left(\mathrm{h}-8\mathrm{a},-3\mathrm{k}\right)$  Also, when  $\mathrm{y}=-3\mathrm{k},\mathrm{x}=\frac{9{\mathrm{k}}^2}{4\mathrm{a}}$