The power of the point  $B\left(-1,1\right)$ with respect to  the circle $S\equiv x^2+y^2-2x-4y+3=0$ is p. If the length of the tangent drawn from B to the circle S = 0 is t, then the  point (2,3) with respect to the circle  $S^{\text{'}}=0$ having centre at  $\left(p,t^2\right)$ and passing through the origin. 

$S\equiv x^2+y^2-2x-4y+3=0,B=(-1,1)$
Power of the point B is$\mathrm{P}\Rightarrow \mathrm{P}=1+1+2-4+3=3$
length of tangent from B to S = 0 is t
$\Rightarrow t=\sqrt{S_{11}}=\sqrt{3}$
$\therefore$ Equation of circle having centre (3, 3) and radius
$\begin{array}{r}\sqrt{18}\text{ is }(x-3{)}^2+(y-3{)}^2=18\\ \Rightarrow S^{\mathrm{\prime }}\equiv x^2+y^2-6x-6y=0\end{array}$
power of (2, 3) = 4 + 9 – 12 – 18 < 0
$\therefore$ point (2, 3) lies inside the circle S' = 0