Two rational points on a circle with irrational centre

Eqn. 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)

If β is the square of a rational no. , h is rational.

Hence h can have max. of 2 rational values