0
$\begingroup$

This question already has an answer here:

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!

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What is the radius, or doesn't that matter? At least include the equation in the question. $\endgroup$ – coffeemath Jun 11 at 16:19
  • $\begingroup$ Try this using general equation form $\endgroup$ – SANJAY KUMAR PANDEY Jun 11 at 17:19
  • $\begingroup$ It is just that the circle doesn't passes through origin $\endgroup$ – SANJAY KUMAR PANDEY Jun 11 at 17:22
  • $\begingroup$ 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. $\endgroup$ – coffeemath Jun 12 at 1:58
0
$\begingroup$

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 $$(x-\pi)^2+(y-e)^2=(\sqrt{\pi^2+e^2})^2 ?$$

$\endgroup$
  • $\begingroup$ Nope but can't you explain using general equation form $\endgroup$ – SANJAY KUMAR PANDEY Jun 11 at 17:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.