Let two points $({\text{x}}_{\text{1}},{\text{ y}}_{\text{1}})$  and  $({\text{x}}_{\text{2}},{\text{ y}}_{\text{2}})$ lie on the circumference of a circle with centre $\left(\text{π}\text{, }\text{e}\right)$ where ${\text{x}}_1\text{, }{\text{y}}_1\text{, }{\text{x}}_2\text{, }{\text{y}}_2\in \text{Q}$

Point $\left(\text{x},\text{y}\right)$ equidistant from $\left({\text{x}}_1,{\text{y}}_1\right)$ & $({\text{x}}_2,{\text{y}}_2)$ will lie on the perpendicular bisector of this line segment.

$\text{y}-\frac{{\text{y}}_1+{\text{y}}_2}{2}=-\left(\frac{\left({\text{x}}_2-{\text{x}}_1\right)}{{\text{y}}_2-{\text{y}}_1}\right)\left(\text{x}-\frac{\left({\text{x}}_1+{\text{x}}_2\right)}{2}\right)$

$\Rightarrow \text{p}\text{x}+\text{q}\text{y}+\text{r}=0$ where $\text{p},\text{q},\text{r}\text{ ∈Q}$

Now it must pass through $\left(\text{π},\text{e}\right)$$\Rightarrow \text{p}\text{π}+\text{q}\text{e}+\text{r}=0$

$\Rightarrow \text{p}=\text{q}=\text{r}=0$

There is no such line possible, hence atmost one point lies on the circle.