If the ratio of the lengths of tangents drawn | vedclass

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(C) The length of the tangent from a point $(x_1, y_1)$ to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 +

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$f^2 + g^2 + 4g + 4f + 2 = 0$

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(C) The length of the tangent from a point $(x_1, y_1)$ to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.For the point $(f, g)$ and the circle $x^2 + y^2 - 6 = 0$,the length of the tangent $L_1 = \sqrt{f^2 + g^2 - 6}$.For the point $(f, g)$ and the circle $x^2 + y^2 + 3x + 3y = 0$,the length of the tangent $L_2 = \sqrt{f^2 + g^2 + 3f + 3g}$.Given the ratio $L_1 : L_2 = 2 : 1$,we have $\frac{L_1}{L_2} = 2$,so $\frac{L_1^2}{L_2^2} = 4$.Substituting the values: $\frac{f^2 + g^2 - 6}{f^2 + g^2 + 3f + 3g} = 4$.$f^2 + g^2 - 6 = 4(f^2 + g^2 + 3f + 3g)$.$f^2 + g^2 - 6 = 4f^2 + 4g^2 + 12f + 12g$.$3f^2 + 3g^2 + 12f + 12g + 6 = 0$.Dividing by $3$,we get $f^2 + g^2 + 4f + 4g + 2 = 0$.

If the ratio of the lengths of tangents drawn from the point $(f, g)$ to the circles $x^2 + y^2 = 6$ and $x^2 + y^2 + 3x + 3y = 0$ is $2 : 1$,then:

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