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(C) 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

Vedclass · Accepted Answer

$\frac{-28}{3}$

Vedclass · Answer

(C) 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 $C_2: 3x^2+3y^2-x-y-\lambda=0$,we first normalize the equation by dividing by $3$:$x^2+y^2-\frac{1}{3}x-\frac{1}{3}y-\frac{\lambda}{3}=0$.The length of the tangent $L_2$ from $(1,2)$ is:$L_2 = \sqrt{1^2+2^2-\frac{1}{3}(1)-\frac{1}{3}(2)-\frac{\lambda}{3}} = \sqrt{1+4-\frac{1}{3}-\frac{2}{3}-\frac{\lambda}{3}} = \sqrt{5-1-\frac{\lambda}{3}} = \sqrt{4-\frac{\lambda}{3}}$.Given the ratio $\frac{L_1}{L_2} = \frac{3}{4}$,we have:$\frac{2}{\sqrt{4-\frac{\lambda}{3}}} = \frac{3}{4} \Rightarrow \sqrt{4-\frac{\lambda}{3}} = \frac{8}{3}$.Squaring both sides:$4-\frac{\lambda}{3} = \frac{64}{9} \Rightarrow \frac{\lambda}{3} = 4-\frac{64}{9} = \frac{36-64}{9} = -\frac{28}{9}$.Multiplying by $3$,we get $\lambda = -\frac{28}{3}$.

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

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