The lengths of the tangents from the point (1 | vedclass

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(D) The length of the tangent from a point $(x_1, y_1)$ to a circle $S: x^2+y^2+2gx+2fy+c=0$ is given by $\sqrt{S_1} = \sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c

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$\frac{21}{4}$

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(D) The length of the tangent from a point $(x_1, y_1)$ to a circle $S: x^2+y^2+2gx+2fy+c=0$ is given by $\sqrt{S_1} = \sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}$.For the first circle $C_1: x^2+y^2+x+y-4=0$,the length of the tangent $L_1$ from $(1,2)$ is:$L_1 = \sqrt{1^2+2^2+1+2-4} = \sqrt{1+4+1+2-4} = \sqrt{4} = 2$.For the second circle $C_2: 3x^2+3y^2-x-y-k=0$,we first normalize it to $x^2+y^2-\frac{1}{3}x-\frac{1}{3}y-\frac{k}{3}=0$.The length of the tangent $L_2$ from $(1,2)$ is:$L_2 = \sqrt{1^2+2^2-\frac{1}{3}(1)-\frac{1}{3}(2)-\frac{k}{3}} = \sqrt{1+4-\frac{1}{3}-\frac{2}{3}-\frac{k}{3}} = \sqrt{5-1-\frac{k}{3}} = \sqrt{4-\frac{k}{3}}$.Given the ratio $\frac{L_1}{L_2} = \frac{4}{3}$,we have:$\frac{2}{\sqrt{4-\frac{k}{3}}} = \frac{4}{3} \Rightarrow \frac{1}{\sqrt{4-\frac{k}{3}}} = \frac{2}{3}$.Squaring both sides:$\frac{1}{4-\frac{k}{3}} = \frac{4}{9} \Rightarrow 4-\frac{k}{3} = \frac{9}{4}$.$\frac{k}{3} = 4 - \frac{9}{4} = \frac{16-9}{4} = \frac{7}{4}$.$k = 3 	imes \frac{7}{4} = \frac{21}{4}$.

The lengths of the tangents from the point $(1,2)$ to the circles $x^2+y^2+x+y-4=0$ and $3x^2+3y^2-x-y-k=0$ are in the ratio $4:3$. Then the value of $k$ is:

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