If x2−xy+y2−4x−4y+16=0  an equation in x and y  has only one real solution, then values of x and y

Given equation isx2−(y+4)x+y2−4y+16=0 Since x is real, D≥0⇒(y+4)2−4y2−4y+16≥0⇒ −3y2+24y−48≥0⇒ y2−8y+16≤0⇒ (y−4)2≤0⇒ y−4=0⇒ y=4 Since equation is symmetric in x and y,x=4 .