The number of pairs of reals (x, y) such that $x = x^{2} + y^{2}$ and $y = 2 x y$ is

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Solution

The correct option is D 4
$x = x^{2} + y^{2}$ & $y = 2 x y$
I when y =0
$\Rightarrow x = x^{2}$
$\Rightarrow x = 0, x = 1$
$∴$ solutions are (0,0) (1,0)
II when $x = \frac{1}{2}$
$\Rightarrow \frac{1}{2} - \frac{1}{4} = y^{2} \Rightarrow \frac{1}{4} = y^{2} \Rightarrow y = \pm \frac{1}{2}$
$∴$ solutions $(\frac{1}{2}, \frac{1}{2}) (\frac{1}{2}, - \frac{1}{2})$

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