A pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve $x^{2} + y^{2} = 4$ with $x + y = a$ . The set containing the value of $a$ is

{- 2, 2}

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{- 3, 3}

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{- 4, 4}

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{- 5, 5}

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Solution

The correct option is A ${- 2, 2}$
To make the given curves $x^{2} + y^{2} = 4$ and $x + y = 9$ homogeneous
$x^{2} + y^{2} - 4 {(\frac{x + y}{a})}^{2} = 0$
$\Rightarrow a^{2} (x^{2} + y^{2}) - 4 (x^{2} + y^{2} + 2 x y) = 0$
$\Rightarrow x^{2} (x^{2} - 4) + y^{2} (a^{2} - 4) - 8 x y = 0$
Since, this is perpendicular pair of straight lines
$∴ a^{2} - 4 + a^{2} = 0 \Rightarrow a^{2} = 4 \Rightarrow a = \pm 2$
Hence the required set of $a$ is $(- 2, 2)$

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