# The number of rationals points on the circumference of a circle having centre $(\pi,e)$ [duplicate]

My main doubt is that when we use the general equation of this particular circle
it is said that $$e = \exp(1)$$ should be rational multiple of $$\pi$$ and so on.
Can you clear why does it occurs!

## marked as duplicate by Wojowu, G Cab, YuiTo Cheng, Shailesh, LeucippusJun 12 at 0:35

• What is the radius, or doesn't that matter? At least include the equation in the question. – coffeemath Jun 11 at 16:19
• Try this using general equation form – SANJAY KUMAR PANDEY Jun 11 at 17:19
• It is just that the circle doesn't passes through origin – SANJAY KUMAR PANDEY Jun 11 at 17:22
• It's not hard to show that, whatever the radius, there cannot be two (or more) rational points on the circumference of a circle centered at $(\pi,e).$ [need to use that these coordinates are rationally independent. – coffeemath Jun 12 at 1:58

COMMENT.-You say nothing about the radius. Let the radius of the circle be equal to $$\sqrt{\pi^2+e^2}$$. The problem is that you assume that there is a rational point in the circumference, which is not evident in any way. Your question could however be interesting in the sense that it is not evident that there is not a rational point. Assuming yes, then we would have equality in the circle of equation $$(x-\pi)^2+(y-e)^2=(\sqrt{\pi^2+e^2})^2$$ supposing $$(a,b)$$ is a rational point in this circle we would have the interesting equality
$$a^2-2a\pi+b^2-2be=0$$
It is a known open problem to know if $$\pi+e$$ is rational, so it could be pertinent to propose the following problem related to yours one:
Is there a rational point in the circle of equation $$(x-\pi)^2+(y-e)^2=(\sqrt{\pi^2+e^2})^2 ?$$