If $x$ and $y$ are real, and $x^{2} + y^{2} - 2 x y + 4 x + 4 y + 12 = 0$ , then $y$ can take the values :

- 1

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Solution

The correct option is A $- 1$
$x^{2} + y^{2} - 2 x y + 4 x + 4 y + 12 = 0$
Forming a quadratic in $x$ assuming $y$ constant

$x^{2} + (4 - 2 y) x + y^{2} + 4 y + 12 = 0$

Applying quadratic formula

$x = \frac{- (4 - 2 y) \pm \sqrt{{(4 - 2 y)}^{2} - 4 (1) (y^{2} + 4 y + 12)}}{2 (1)} x = \frac{(2 y - 4) \pm \sqrt{16 + 4 y^{2} - 16 y - 4 y^{2} - 16 y - 48}}{2} x = \frac{(2 y - 4) \pm \sqrt{- 32 y - 32}}{2}$

Now $y$ and $x$ are both real , therefore the term under square root must be positive or zero

$\Rightarrow - 32 y - 32 \geq 0 \Rightarrow 32 y + 32 \leq 0 \Rightarrow y \leq - 1$

So, option A is correct.

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