Given x^2 - xy + y^2 = 4(x+y-4) where x, y both are real then the number of pairs (x, y) satisfying the equation will be
(A) Only one
(D) No pair
The Attempt at a Solution
The equation can be written as (x-2)^2 + (y-2)^2 + 8 = xy on opening the brackets and completing the square.
Beyond this, I really don't know what to do. Hit and trial doesn't seem to work very well either.