A circle in the plane contains at most four lattice points?

Let $\cal C$ be a circle in ${\mathbb R}^2$ : $\cal C=\lbrace (x,y)\in{\mathbb R}^2 | (x-x_0)^2+(y-y_0)^2=r^2\rbrace$ for some constants $x_0,y_0,r$.

What is the maximal number of points that can be contained in ${\cal C}\cap {\mathbb Z}^2$ ? I conjecture it is $4$, attained for the "trivial" case $x_0=y_0=0,r=1$.

The unit circle centred on the origin can be parametrised by $$x=\frac {1-t^2}{1+t^2}; y=\frac {2t}{1+t^2}$$

Any rational value of $t$ gives rational values of $x$ and $y$. This can be scaled by a factor $r$ to give a circle of radius $r$.

Choose $n$ such points, and then choose a radius which clears all the denominators - the resulting circle will have at least $n$ integer points.

There is no upper bound. On the circle:

$$ x^2+y^2 = 5^k $$ there are exactly $4k+4$ lattice points. That follows from the fact that the number of representations of $n$ as $x^2+y^2$ is given by four times a multiplicative function that depends on the number of divisors of the form $4k+1$ and the number of divisors of the form $4k+3$.

There is no upper bound. Take the product of, for example, several consecutive primes that are $1 \pmod 4,$ and let $r = 5 \cdot 13 \cdot 17 \cdot 29.$ Find all Pythagorean triples with that hypoteneuse $r.$ Those all occur as points on the circle $x^2 + y^2 = r^2$