Let two points (x1,y1) and (x2,y2) lie on the circumference of a circle with centre (π,e) where x1,y1,x2,y2∈Q
Point (x,y) equidistant from (x1,y1) & (x2,y2) will lie on the perpendicular bisector of this line segment. y−2y1+y2=−(y2−y1(x2−x1))(x−2(x1+x2)) ⇒px+qy=r=0 where p,q,r∈Q
Now it must pass through (π,e)⇒pπ+qe+r=0 ⇒p=q=r=0
There is no such line possible, hence at most one point lies on the circle.