The number of pairs (x , y) which will satisfy the equation
$x^{2} - x y + y^{2} = 4 (x + y - 4),$ is

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None of these

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Solution

The correct option is A 1
$x^{2} - x y + y^{2} = 4 (x + y - 4)$
$⟹ x^{2} - x (y + 4) + y^{2} - 4 y + 16 = 0$
If it has real solution then $D \geq 0$
So,
$(y + 4)^{2} - 4 (y^{2} - 4 y + 16) \geq 0$
Or $(y - 4)^{2} \leq 0$
It is only possible when $y = 4$
So for $y = 4$ , there is only one value of $x = 4$ possible
So only one pair $(4, 4)$ is the solution.

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