x2−xy+y2−4x−4y+16=0 represents

Given equation is x2−(y+4)x+y2−4y+16=0Since x is real, we haveD≥0⇒ (y+4)2−4y2−4y+16≥0or   −3y2+24y−48≥0or  y2−8y+16≤0or  (y−4)2≤0or  y - 4 = 0or  y = 4Since the equation is symmetric in r and y, we have x = 4 only.