Can a Circle with an Irrational Center Have More Than Two Rational Points?

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A circle with an irrational center can have at most two rational points. This conclusion is derived from the equations of the circle, specifically the forms (X - x_1)^2 + (Y - y_1)^2 = R^2 and (X - x_2)^2 + (Y - y_2)^2 = R^2. By eliminating R and forming simultaneous linear equations, it is established that for X and Y to remain rational, the rational points must either be identical or collinear.

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vishal007win
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How many rational points can be there on a circle which has an irrational centre?
(rational point is a point which have both x,y as rational numbers)

how to proceed??
answer is: atmost 2
 
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You say a "rational point" has both x and y rational numbers. Is an "irrational point", then, a point that has either x or y or both irrational?
 
I'd think that if a rational point has both x,y rational, an irrational point would be
a point that is not ratinal, so has at least one of x,y irratinal?

write down the equations of the circle for two points

(X - x_1)^2 + (Y - y_1)^2 = R^2
(X - x_2)^2 + (Y - y_2)^2 = R^2

You can eliminate R from them, and write them so X and Y become separated

2 X (x_2 - x_1) + 2 Y (y_2 - y1) = x_2^2 + y_2^2 - x_1^2 - y_1^2

Note that all numbers, except for X and Y are rational. With 2 points it's still
possible to have X or Y irrational. Now add a third point

(X - x_3)^2 + (Y - y_3)^2 = R^2

and combine this equation with the one for the first point producing

2 X (x_3 - x_1) + 2 Y (y_3 - y1) = x_3^2 + y_3^2 - x_1^2 - y_1^2

we now get 2 simultaneous linear equations for X and Y, and it is possible to prove that X and Y cannot be rational unless 2 of the rational points on the circle are identical, or the rational points on the circle lie on a straight line.
 

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