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

0

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Solution

The correct options are
B $- 1$
C $- 3$
D $- 2$
$x^{2} + y^{2} - 2 x y + 4 x + 4 y + 12 = 0$

Forming a quadratic in $y$ assuming $x$ constant

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

Applying quadratic formula, we have

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

Now $x$ and $y$ are both real so the term under the root must be greater than or equal to zero.

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

So, options B,C and D are correct.

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