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 $4 (3-0) = 4 \cdot 3 = 12.$