If the length of the tangent drawn from $(α, β)$ to the circle $x^{2} + y^{2} = 6$ be twice the length of the tangent from the same point to the circle $x^{2} + y^{2} + 3 x + 3 y = 0$ , then prove that $α^{2} + β^{2} + 4 α + 4 β + 2 = 0$ .

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Solution

Length of tangent from point $(α, β)$ to a circle $c = x^{2} + y^{2} + 2 g + 2 f y + c = 0$ equals: $\sqrt{α^{2} + β^{2} + 2 α g + 2 α f + c}$
$c_{1} = x^{2} + y^{2} = 6$
$c_{2} = x^{2} + y^{2} + 3 x + 3 y = 0$
Tangent length for $c_{1} = \sqrt{α^{2} + β^{2} - 6}$
Tangent length for $c_{2} = \sqrt{α^{2} + β^{2} + 3 α + 3 β}$
According to the question,
$\sqrt{α^{2} + β^{2} - 6} = 2 \sqrt{α^{2} + β^{2} + 3 α + 3 β}$
$∴ α^{2} + β^{2} - 6 = 4 (α^{2} + β^{2} + 3 α + 3 β)$
$∴ 0 = 3 α^{2} + 3 β^{2} + 12 α + 12 β + 6$
$∴ 0 = α^{2} + β^{2} + 4 α + 4 β + 2$

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