From any point on the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ tangents are drawn to the circle $x^{2} + y^{2} + 2 g x + 2 f y + c \sin^{2} α + (g^{2} + f^{2}) \cos^{2} α = 0$ then angle between tangents is

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$0^{0}$

$α$

$2 α$

$\frac{π}{2}$

answer is B.

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Detailed Solution

$x^{2} + y^{2} + 2 g y + 2 f y + c = 0 L e t (h, k) b e a n y p o i n t o n a b o v e c i r c l e t h e n h^{2} + k^{2} + 2 g h + 2 f k = - c \to (1) l e n g t h o f \tan g e n t f r o m (h, k) t o s e c o n d c i r c l e = \sqrt{h^{2} + k^{2} + 2 g h + 2 f k + c \sin^{2} α + (g^{2} + f^{2}) \cos^{2} α} = \sqrt{(g^{2} + f^{2}) \cos^{2} α - c (1 - \sin^{2} α)} = \sqrt{(g^{2} + f^{2} - c) \cos^{2} α} = \sqrt{(g^{2} + f^{2} - c)} \cos α r^{2} = g^{2} + f^{2} - (c \sin^{2} α + (g^{2} + f^{2}) \cos^{2} α) = (g^{2} + f^{2} - c) \sin^{2} α \Rightarrow r = \sqrt{(g^{2} + f^{2} - c)} \sin α N o w \tan \frac{θ}{2} = \frac{r}{\sqrt{S_{11}}} = \frac{\sqrt{(g^{2} + f^{2} - c)} \sin α}{\sqrt{(g^{2} + f^{2} - c)} \cos α} \Rightarrow \tan \frac{θ}{2} = \tan α \Rightarrow \frac{θ}{2} = α \Rightarrow θ = 2 α$