If the length of the tangent from $(f, g)$ to the circle $x^{2} + y^{2} = 6$ be twice the length of the tangent from $(f, g)$ to the circle $x^{2} + y^{2} + 3 x + 3 y = 0$ , then will $f^{2} + g^{2} + 4 f + 4 g + 2 = 0$

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Solution

We know that,
Length of Tangent from a Point (a,b) to a Circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
is $L = \sqrt{a^{2} + b^{2} + 2 g a + 2 f b + c}$

So, According to Question
$\sqrt{f^{2} + g^{2} - 6} = 2 \sqrt{f^{2} + g^{2} + 3 f + 3 g} \Rightarrow f^{2} + g^{2} - 6 = 4 (f^{2} + g^{2} + 3 f + 3 g) \Rightarrow 3 f^{2} + 3 g^{2} + 12 f + 12 g + 6 = 0 \Rightarrow f^{2} + g^{2} + 4 f + 4 g + 2 = 0$

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(f, g)

x^{2} + y^{2} = 6

x^{2} + y^{2} + 3 x + 3 y = 0

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x^{2} + y^{2} = 6

x^{2} + y^{2} + 3 x + 3 y = 0

α^{2} + β^{2} + 4 α + 4 β + 2 = 0

x^{2} + y^{2} - 3 x - 4 y + 2 = 0

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

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\frac{π}{2},

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

x^{2} + y^{2} + 2 g x + 2 f y + c {sin}^{2} α + (g^{2} + f^{2}) {cos}^{2} α = 0

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