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Let $\mathrm{S}$ be a circle concentric with the circle $3 x^2+3 y^2+x+y-1=0$. If the length of the tangent drawn from a point $(2,-2)$ to the given circle is the radius of the circle S, then the power of the point $(2,1)$ with respect to the circle $\mathrm{S}$ is
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Verified Answer
The correct answer is:
$\frac{-29}{18}$
Given circle is $3 x^2+3 y^2+x+y-1=0$
$$
\begin{aligned}
& \Rightarrow x^2+y^2+\frac{1}{3} x+\frac{1}{3} y-\frac{1}{3}=0 \\
& \therefore \text { Centre }=\left(-\frac{1}{6},-\frac{1}{6}\right), \text { radius }=\frac{\sqrt{14}}{6}
\end{aligned}
$$
Length of tangents $P A$,
$$
\begin{aligned}
& r_1=\sqrt{O P^2-O A^2} \\
& =\sqrt{\left(2+\frac{1}{6}\right)^2+\left(-2+\frac{1}{6}\right)^2-\frac{14}{36}}=\sqrt{\frac{23}{3}}
\end{aligned}
$$
$\therefore$ Equation of required circle $S$ is
$$
\left(x+\frac{1}{6}\right)^2+\left(y+\frac{1}{6}\right)^2=\frac{23}{3}
$$
The power of the point $(2,1)=S_1$
$$
=\left(2+\frac{1}{6}\right)^2+\left(1+\frac{1}{6}\right)^2-\frac{23}{3}=-\frac{29}{18}
$$
$$
\begin{aligned}
& \Rightarrow x^2+y^2+\frac{1}{3} x+\frac{1}{3} y-\frac{1}{3}=0 \\
& \therefore \text { Centre }=\left(-\frac{1}{6},-\frac{1}{6}\right), \text { radius }=\frac{\sqrt{14}}{6}
\end{aligned}
$$
Length of tangents $P A$,
$$
\begin{aligned}
& r_1=\sqrt{O P^2-O A^2} \\
& =\sqrt{\left(2+\frac{1}{6}\right)^2+\left(-2+\frac{1}{6}\right)^2-\frac{14}{36}}=\sqrt{\frac{23}{3}}
\end{aligned}
$$
$\therefore$ Equation of required circle $S$ is
$$
\left(x+\frac{1}{6}\right)^2+\left(y+\frac{1}{6}\right)^2=\frac{23}{3}
$$
The power of the point $(2,1)=S_1$
$$
=\left(2+\frac{1}{6}\right)^2+\left(1+\frac{1}{6}\right)^2-\frac{23}{3}=-\frac{29}{18}
$$
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