On the Diophantine Equation $(x-h)^2+(y-k)^2=c$

Question

I am just curious about the equation of the circle centered at (h,k) whose form is we know $(x-h)^2+(y-k)^2=r^2$.

If we consider its solution over the set of integers then we have a Diophantine Equation. If we consider the unit circle $x^2+y^2=1$ we know that it has the solutions {(0,1),(1,0),(-1,0),(0,-1)} am I correct if I say that Diophantine Equation of this form has at most four solutions?

And what would be the relationships among the variables i.e. h, k and r such that the given equation will have one, two, three or four solutions?

Thank you in advance.

Answer 1

It is easy to argue that the equation has finitely many solutions: any solution must satisfy $h-r \leq x \leq h+r$ and $k+r \leq y \leq k+r$.

Now, the number of points depends on $r$. For each fixed $r$, the number of solutions is easily obtained from the number of (possibly trivial, not necessarily positive or primitive) Pytagorean triple $(a,b,r)$.

Therefore, for each way of writing the given $r$ as $r =l(m^2+n^2)$ with $gcd(m,n)=1$ and $m,n$ of opposite parity we get the following points:

$$x=h \pm 2lmn \\ y=k \pm l(m^2-n^2)$$ and $$x=h \pm l(m^2-n^2) \\ y=k \pm 2lmn$$

You can then build equations with as many solutions as you want, by making sure that $r$ is divisible by as many terms of the form $m^2+n^2$ as you want (i.e. divisible by many different "hypothenuses" from primitive pytagorean triples).

As for fewer solutions, you can never get less than $4$: $(h-r,0) ; (h+r,0), (0, k-r) ; (0,k+r)$ are always four distinct solutions for $r>0$.

Answer 2

As André Nicolas said, there can be arbitrary many solutions. Indeed, this holds already when the circle is centered around $0$. In that case, you know that there are infinitely many rational solutions $p_i = (x_{i,1},x_{i,2})$ to the equation $x^2+y^2=1.$ So, take $m$ rational solutions, $p_1, \ldots, p_m,$ distinct from the four integer solutions you know of. Take the lcm of the denominators in all $$(x_{1,1},x_{1,2},x_{2,1},x_{2,2} \ldots, x_{m,1},x_{m,2})$$ and consider a circle of radius $n$ centered at the origin. This circle will then have more than $m$ integer solutions.