Let's solve this problem using substitution. Given that x-y=8, x = 8 + y.
(Then x^2 = 64 + 16y + y^2)
This other equation is (x-2)^2 + (y-1)^2 = 25.
Easier to substitute 8 + y for x in (x-2)^2:
(8 + y - 2)^2 + (y-1)^2 = 25
(6 + y)^2 + (y-1)^2 = 25
36 + 12y + y^2 + y^2 - 2y + 1 = 25
Re-writing this in descending powers of y:
2y^2 + 10y + 36 + 1 = 25
Then 2y^2 + 10y - 12 = 0
Reduce by division by 2: y^2 + 5y - 6 = 0 = (y+3)(y+2) = 0
Then y=-3 and y=-2. From each of these we get x: x = 8 + y
So x = 8 - 3 = 5 and x = 8 - 2 = 6. There are common solutions.
Try (5, -3) and (6, -2). Do these points satisfy both of the given equations? If they do, you've shown that we have common solutions.