Wolfram Alpha will provide integer solutions to arbitrary circle equations. I'm trying to understand how it's able to calculate them, but despite a fair bit of digging I haven't found any discussion of how to get either the number of, or which, integer solutions to a given circle. Plenty of discussion of lattice points inside a circle, related to the Gauss circle problem, and some discussion of circles centered on the origin, but nothing for the general case.
Wolfram Alpha can quickly determine there are $12$ integer solutions to the circle $x^2-10 (x+y)+y^2+50 = 50$ - how?
As long as the coefficients of $x^2, \; y^2$ are $1$ and the coefficients of $x,y$ are even, this is quite easy. What you get by completing two squares is $$ (x-5)^2 + (y-5)^2 = 50. $$ Fine, so define new variables, $$ u = x-5, \; \; \; v = y - 5, $$ and count the (integer pair) solutions to $$ u^2 + v^2 = 50. $$ For each pair, return by $x = u + 5, \; \; y = v + 5.$ It is easy enough to plot these.
However, what if you had some very large target number $n$ and had to count the number of integer pair solutions to $$ u^2 + v^2 = n? $$ Well, if you can factor $n,$ you can make a complete list of all numbers that divide it, including $1$ and $n$ itself. Ignore the even divisors. Count the number of divisors that leave a remainder of 1 when divided by 4, call that count $C_1.$ Put another way, this is the count of $d > 0, \; \; d | n, \; \; d \equiv 1 \pmod 4.$ Next, count the number of divisors that leave a remainder of 3 when divided by 4, call that count $C_3.$ This is the count of $d > 0, \; \; d | n, \; \; d \equiv 3 \pmod 4.$ The number of integer lattice points on the circle is $$ 4 (C_1 - C_3).$$ For $n = 50,$ the divisors are $1,2,5,10,25,50.$ So $C_1 = 3$ and $C_3 = 0,$ and the number of integer points is $4 (3-0) = 4 \cdot 3 = 12.$