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The power of the point $B(-1,1)$ with respect to the circle $S \equiv x^2+y^2-2 x-4 y+3=0$ is $p$. If the length of the tangent drawn from $B$ to the circles $S=0$ is $t$, then the point $(2,3)$ with respect to the circle $S^{\prime}=0$ having centre at $\left(p, t^2\right)$ and passing through the origin.
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Verified Answer
The correct answer is:
lies inside the circle S' = 0
Given equation of circle
$$
\begin{array}{rlrl}
& & S & \equiv x^2+y^2-2 x-4 y+3=0 \\
\therefore & \quad p & =(-1)^2+(1)^2-2(-1)-4(1)+3 \\
& & =1+1+2-4+3=3 \\
\because & & t & =\sqrt{p} \Rightarrow t=\sqrt{3}
\end{array}
$$
Now, circle whose centre is $\left(p, t^2\right)$, i.e. $(3,3)$
$$
(x-3)^2+(y-3)^2=r^2
$$
Since, this circle passes through $(0,0)$
$$
\begin{array}{ll}
\therefore & (0-3)^2+(0-3)^2=r^2 \\
\Rightarrow & r^2=9+9=18
\end{array}
$$
So, circle $S^{\prime}$ will be
$$
(x-3)^2+(y-3)^2=18
$$
Now, point $(2,3)$ w.r.t. to circle
$$
\begin{aligned}
& (x-3)^2+(y-3)^2=18 \text { is } \\
& =(2-3)^2+(3-3)^2-18 \\
& =1-18=-17 < 0
\end{aligned}
$$
So, point $(2,3)$ lies inside the circle $S^{\prime}=0$
$$
\begin{array}{rlrl}
& & S & \equiv x^2+y^2-2 x-4 y+3=0 \\
\therefore & \quad p & =(-1)^2+(1)^2-2(-1)-4(1)+3 \\
& & =1+1+2-4+3=3 \\
\because & & t & =\sqrt{p} \Rightarrow t=\sqrt{3}
\end{array}
$$
Now, circle whose centre is $\left(p, t^2\right)$, i.e. $(3,3)$
$$
(x-3)^2+(y-3)^2=r^2
$$
Since, this circle passes through $(0,0)$
$$
\begin{array}{ll}
\therefore & (0-3)^2+(0-3)^2=r^2 \\
\Rightarrow & r^2=9+9=18
\end{array}
$$
So, circle $S^{\prime}$ will be
$$
(x-3)^2+(y-3)^2=18
$$
Now, point $(2,3)$ w.r.t. to circle
$$
\begin{aligned}
& (x-3)^2+(y-3)^2=18 \text { is } \\
& =(2-3)^2+(3-3)^2-18 \\
& =1-18=-17 < 0
\end{aligned}
$$
So, point $(2,3)$ lies inside the circle $S^{\prime}=0$
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