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Q.

If the angle between the circles $x^{2} + y^{2} + 4 x - 5 = 0$ and $x^{2} + y^{2} + 2 λ y - 4 = 0$ is $\frac{π}{3}$ , then $λ$ =

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a

$\pm 2$

b

$\pm \sqrt{3}$

c

$\pm \sqrt{6}$

d

$\pm \sqrt{5}$

answer is A.

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Detailed Solution

$\begin{aligned} x^{2} + y^{2} + 4 x - 5 = 0, x^{2} + y^{2} + 2 λ y - 4 = 0 \\ \cos \frac{π}{3} = \frac{- 5 - 4 - 0 - 0}{2 \sqrt{4 + 5} \sqrt{λ^{2} + 4}} \\ \Rightarrow \frac{1}{2} = \frac{- 9}{2 (3) \sqrt{λ^{2} + 4}} \\ \Rightarrow (λ^{2} + 4) = 9 \\ \Rightarrow λ^{2} = 5 \end{aligned}$
$\Rightarrow λ = \pm \sqrt{5}$

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