If the slope of the tangent to the circle $S\equiv x^2+y^2-13=0$ at $\left(2, 3\right)$ is m, then the point $\left(m, \frac{-1}{m}\right)$ is 

$$\begin{gathered} \mathrm{equation} \mathrm{of} \mathrm{tangent} \mathrm{at} (2,3) \mathrm{to} \mathrm{the} \mathrm{circle} \mathrm{S}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-13=0 \mathrm{is} \\ {\mathrm{S}}_1=\mathrm{O} \\ \Rightarrow 2\mathrm{x}+3\mathrm{y}-13=\mathrm{O} \\ \mathrm{Slope}\left(m\right)=\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \\ Now P\left(\mathrm{m}, \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.\right)=\left(\frac{-2}{3}, \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right) \\ Now sub P in circle \\ {S}_{11}=\frac{4}{9}+\frac{9}{4}-13<0 \\ So point P lies inside the circle \end{gathered}$$