If the ratio of the lengths of the tangents d | 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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$-21/4$

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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 first circle $x^2 + y^2 + x + y - 4 = 0$,the length $T_1 = \sqrt{1^2 + 2^2 + 1 + 2 - 4} = \sqrt{1 + 4 + 1 + 2 - 4} = \sqrt{4} = 2$.For the second circle,we first normalize the equation: $x^2 + y^2 - \frac{1}{3}x - \frac{1}{3}y + \frac{k}{3} = 0$.The length $T_2 = \sqrt{1^2 + 2^2 - \frac{1}{3}(1) - \frac{1}{3}(2) + \frac{k}{3}} = \sqrt{1 + 4 - \frac{1}{3} - \frac{2}{3} + \frac{k}{3}} = \sqrt{5 - 1 + \frac{k}{3}} = \sqrt{4 + \frac{k}{3}}$.Given the ratio $\frac{T_1}{T_2} = \frac{4}{3}$,we have $\frac{2}{\sqrt{4 + k/3}} = \frac{4}{3}$.Squaring both sides: $\frac{4}{4 + k/3} = \frac{16}{9} \Rightarrow 36 = 64 + \frac{16k}{3}$.$36 - 64 = \frac{16k}{3}$ $\Rightarrow -28 = \frac{16k}{3}$ $\Rightarrow k = \frac{-28 	imes 3}{16} = \frac{-7 	imes 3}{4} = -\frac{21}{4}$.

If the ratio of the lengths of the tangents drawn from the point $(1, 2)$ to the circles $x^2 + y^2 + x + y - 4 = 0$ and $3x^2 + 3y^2 - x - y + k = 0$ is $4 : 3$,then $k = \dots$

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