If the lengths of the tangents drawn from the point $(1,2)$ to the circles $$ x^2+y^2+x+y-4=0 $$ and $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 $3 x^2+3 y^2-x-y-\lambda=0$ are in the ratio $3: 4$, then $\lambda$ is equal to
$\frac{23}{4}$
$\frac{17}{4}$
$\frac{-28}{3}$
$\frac{19}{4}$
Solution
$
\begin{aligned}
& \text { } C_1: x^2+y^2+x+y-4=0 \\
& \quad C_2: 3 x^2+3 y^2-x-y-\lambda=0 \\
& \Rightarrow \quad x^2+y^2-\frac{x}{3}-\frac{y}{3}-\frac{\lambda}{3}=0
\end{aligned}
$
$\therefore$ Length of tangent drawn by an external point is $\sqrt{S_1}$.
$\therefore$ Length of tangent drawn from $(1,2)$ to $C_1$ and $C_2$ are $L_1$ and $L_2$ respectively.
$
\begin{aligned}
L_1 & =\sqrt{S_1}=\sqrt{1^2+2^2+1+2-4} \\
& =\sqrt{1+4+1+2-4}=\sqrt{4} \\
L_1 & =2
\end{aligned}
$
$
\therefore L_2=\sqrt{S_1^{\prime}}=\sqrt{1^2+2^2-\frac{1}{3}-\frac{2}{3}-\frac{\lambda}{3}}=\sqrt{4-\frac{\lambda}{3}}
$
Given that,
$
\frac{L_1}{L_2}=\frac{3}{4}
$
$\begin{aligned} &=\frac{2}{\sqrt{4-\frac{\lambda}{3}}}=\frac{3}{4} \Rightarrow \sqrt[3]{4-\frac{\lambda}{3}}=8 \\ & \Rightarrow \quad 9\left(4-\frac{\lambda}{3}\right)=64 \Rightarrow 4-\frac{\lambda}{3}=\frac{64}{9} \\ & \Rightarrow \quad 4-\frac{64}{9}=\frac{\lambda}{3} \Rightarrow \frac{36-64}{9}=\frac{\lambda}{3} \\ & \lambda=-\frac{28}{3}\end{aligned}$