1Eqn. of the circle : x2 + (y-√2)2 = a2
where a is the radius.
Let (h,k) be a pt. on the circle.
hence h2+k2-2√2k+2=a2
or, h2+k2-2√2k=a2-2=β (say)
If β is rational then comparing, we get h2+k2=β
But 2√2k = 0.
Hence h2=β or, h=±√β
If β is irrational, h2+k2=0 hence k=0 (as both the terms are squares hence each of them must be 0)
bit 2√2k=β and again k=0 which is a contradiction.
Hence at the most there can be 2 rational points on the circle (h=±√β)
