Agfdhyk

What is the number of rational points on a circle having centre as $(pi,e)$?

At most $2$: if there were three points with rational coordinates, the center would have rational coordinates, too.

Proof: let $A$, $B$ and $C$ be points on a circle with rational coordinates. The center $M$ of the circle passing through these points is the intersection of the perpendicular bisectors of $AB$ and $AC$. Now the equations of those bisectors contain only rational coefficients.

Solving that linear system, Cramer’s rule, for example, reveals that their intersection has rational coefficients, too.

Addendum: There are circles with irrational coordinates of the centre, which have two rational points on it. Take two points with rational coordinates; the equation of their perpendicular bisector has rational coefficients, hence surely there are irrational points on that bisector.

Example:
$$(x-1/2-sqrt2)^2+(y-1/2+sqrt2)^2=(1/2+sqrt2)^2
+(1/2-sqrt2)^2.$$
Obviously $(0,0)$ and $(1,1)$ belong to that circle.

As comments imply, it is possible to have a point on the circle, and the other answer implies we can't have three. Let me complement this by saying that existence of a circle with this center with two rational points is an open problem.

Indeed, two points $P,Q$ lie on some circle centered at $(e,pi)$ iff this point lies on the perpendicular bisector of the interval $PQ$. This line has an equation with rational coefficients, and conversely every line with rational coefficients is a bisector of some segment with rational endpoints. Therefore two points can lie on such a circle iff we can have a rational solution to the equation $ae+bpi+c=0$. This, as you can imagine, is widely believed to not be possible, but it's still open - indeed, we don't even know whether $e/pi$ is rational, and if it were, we could even have a relation above with $c=0$.

There is at most one rational point on that circle, assuming $pi, e$ and $1$ are linearly independent over $Bbb Q$ (which seems to be an open problem).

A rational point $(p, q)$ on the circle means that there is a real number $r$ such that
$$
(p-pi)^2 + (q-e)^2 = r^2
$$
If there is another rational point $(s, t)$ on that same circle, then we also have
$$
(s-pi)^2 + (t-e)^2 = r^2
$$
giving us
$$
(p-pi)^2 + (q-e)^2 =(s-pi)^2 + (t-e)^2\
underbrace{(2s-2p)pi}_{text{rational multiple of }pi} + underbrace{(2t-2q)e}_{text{rational multiple of }e} + underbrace{p^2 - s^2 + q^2 - t^2}_{text{rational multiple of }1} = 0
$$
That the two points are distinct implies the three braced terms are not all zero, meaning $pi, e$ and $1$ are linearly dependent over $Bbb Q$.