The angle between the lines joining the origin to the point of intersection of $l x + m y = 1 and x^{2} + y^{2} = a^{2}$ is

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$\frac{π}{4}$

$\frac{π}{2}$

$\cos^{- 1} (\frac{1}{a \sqrt{l^{2} + m^{2}}})$

$2 \cos^{- 1} (\frac{1}{a \sqrt{l^{2} + m^{2}}})$

answer is D.

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Detailed Solution

$E q u a t i o n o f p a i r o f l i n e s p a s \sin g t h r o u g h p o i n t o f i n t e r s e c t i o n o f g i v e n l i n e s a r e x^{2} + y^{2} - a^{2} {(l x + m y)}^{2} = 0 \Rightarrow x^{2} + y^{2} - a^{2} (l^{2} x^{2} + m^{2} y^{2} + 2 l m x y) = 0 \Rightarrow (1 - a^{2} l^{2}) x^{2} - 2 a^{2} l m x y + (1 - a^{2} m^{2}) y^{2} = 0 N o w \cos θ = \frac{2 - a^{2} (l^{2} + m^{2})}{\sqrt{a^{4} (l^{2}} - m^{2})^{2} + 4 a^{4} l^{2} m^{2}} = \frac{2 - a^{2} (l^{2} + m^{2})}{a^{2} (l^{2} + m^{2})} \cos θ = \frac{2}{a^{2} (l^{2} + m^{2})} - 1 \Rightarrow 1 + \cos θ = \frac{2}{a^{2} (l^{2} + m^{2})} \Rightarrow 2 \cos^{2} (\frac{θ}{2}) = \frac{2}{a^{2} (l^{2} + m^{2})} \Rightarrow \cos (\frac{θ}{2}) = \frac{1}{a^{2} (l^{2} + m^{2})} \Rightarrow θ = 2 \cos^{- 1} (\frac{1}{a^{2} (l^{2} + m^{2})})$