rational pts

is it only one??...i m not sure[3]

yeah thats right,answer is one,but can you post ur proof

(x-Ï€)²+(y-e)²=a²
x²-2Ï€x+Ï€²+y²-2ey+e²=a²

when a²=Ï€²+e²

then it passes through (0,0)

this not a proof i think[3] may b just a hint...

abhirup has shown that there exist circles that have (\pi,e) as centre and have at least one rational point [both x and y coordinates are rational numbers]

We will now prove that there exists at most one such point.

Suppose having found a point (x_1,y_1) that is a rational point on this circle. Now, suppose there exists another point (x_2,y_2) that is also rational.

The equation of the circle is (x-\pi)^2+(y-e)^2 = r^2, so that

(x_1-\pi)^2+(y_1-e)^2 = (x_2-\pi)^2+(y_2-e)^2

from which we obtainx_1^2-x_2^2 + y_1^2-y_2^2 = 2[\pi(x_1-x_2)+e(y_1-y_2)]

Since LHS is rational we must have \pi(x_1-x_2)+e(y_1-y_2) = 0

Again, if x_1 \ne x_2; y_1 \ne y_2, we obtain \frac{\pi}{e} to be rational [which according to the authors of the problem must be a contradiction. I say this because i am not sure how to conclude that \frac{\pi}{e} is irrational, though a rational human would assume so :D]

(...contd). So that means we must have x_1 = x_2 and y_1 = y_2 and that means at most one rational point exists.