# Thread: Rational points on a circle with centre whose coordinates are (pi,e)

1. ## Rational points on a circle with centre whose coordinates are (pi,e)

Here is the question: -

There is a circle with centre (pi,e). The radius of teh circle could be anything. How many rational points (points whose both coordinates are rational) exist on the circle?

I have been unable to come p with any line of thought as of now. Please contribute!

2. Originally Posted by dharavsolanki
Here is the question: -

There is a circle with centre (pi,e). The radius of teh circle could be anything. How many rational points (points whose both coordinates are rational) exist on the circle?

I have been unable to come p with any line of thought as of now. Please contribute!
You know there is at least 1.
Select ANY rational numbers (x,y) to be a point on the circle.

The question now is: Does there exist another point rational point that is r distance from $(\pi \text{,} e)$

$r^2 = (x-\pi)^2 + (y-e)^2$

$r^2 = x^2 - 2x\pi +\pi^2 + y^2 -2ye + e^2$

$r^2 -\pi^2 -e^2 = x^2 + y^2 -2(x\pi+ye)$

$r^2 -\pi^2 -e^2 + 2(x\pi+ye) = x^2 + y^2$

Can you find a 2nd rational coordinate pair to give the same result?

.

3. How do I know there is atleast one point on the circle which says so?

4. Originally Posted by dharavsolanki
There is a circle with centre (pi,e). The radius of the circle could be anything. How many rational points (points whose both coordinates are rational) exist on the circle? I have been unable to come p with any line of thought as of now.
Originally Posted by dharavsolanki
How do I know there is atleast one point on the circle which says so?
This question is so poorly written as to make answering it impossible.
Are you asking “given any radius $r>0$ then is there a point on the circle centered at $(\pi,e)$ and radius $r$ is there a point on that circle that has rational coordinates?”

On the other hand, you were shown that given any point with rational coordinates say $(a,b)$ there is always a circle centered at $(\pi,e)$ containing $(a,b)$: $(x-\pi)^2+(y-e)^2=(a-\pi)^2+(b-e)^2$.

Please try to clarify the wording of this question.
What exactly are you asking? What does “to come p with any line of thought as of now” mean?

5. Originally Posted by Plato
What exactly are you asking? What does “to come p with any line of thought as of now” mean?
Are you poking on a typo, sir? I meant unable to come up with any line of thought.

Originally Posted by Plato
Are you asking “given any radius then is there a point on the circle centered at and radius is there a point on that circle that has rational coordinates?”
Yes. Exactly! That is the question.
The first reply to the question makes the argument that we already know there's one rational point, reducing the entire procedure to asking is there any other such point?

Here lies my problem. How do we know there is atleast one rational point on the given circle?

6. Originally Posted by dharavsolanki
How do we know there is atleast one rational point on the given circle?
Because the length of the radius can be any value, we can select ANY point and call it a point on the circle.
The distance between the two points defines the RADIUS and thus the circle.

The center of the circle is $(\pi,e)$

A random point (call it P) has coordinates $\left( \dfrac{59}{3} \text{,} \dfrac{113}{4}\right)$

Is it possible for you to calculate the distance between the center of the circle and point P?

If you can calculate the distance, you have found the radius of the circle and thus proof that the point is on the circle.

An 8th grade student was able to do this:

Radius = $\sqrt{\left( \dfrac{59-3\pi}{3}\right)^2 + \left(\dfrac{113-4e}{4}\right)^2 }$

.
If you intended to ask: with the given irrational coordinates for the center of the circle, and some rational points on the circle, is it possible to have a RATIONAL RADIUS?
And you can prove that along the same line of reasoning that the square root of 2 is not rational.

,

,

,

,

,

### the number of rational points on circle having center(Ï€,e)

Click on a term to search for related topics.