From a point P on the line 4x–3y =6 two tangents are drawn to the circle $x^2+y^2-6x-4y+4=0$ . If the angle between these tangents is ${\tan }^{-1}\left(\frac{{\displaystyle 24}}{{\displaystyle 7}}\right)$, then P =

$$\begin{gathered} \mathrm{Let} P(\mathrm{x},\mathrm{y}) \mathrm{be} \mathrm{a} \mathrm{point} \mathrm{in} 4\mathrm{x}-37=6 \\ C=(3,2) \\ \mathrm{r}=\sqrt{9+4-4}=3 \\ Given \tan \theta =\frac{24}{7} \\ \Rightarrow \tan \left(\raisebox{1ex}{$\mathrm{\theta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=\frac{3}{4} \\ \tan \left(\raisebox{1ex}{$\mathrm{\theta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=\raisebox{1ex}{$\mathrm{r}$}\!\left/ \!\raisebox{-1ex}{$\sqrt{{\mathrm{S}}_{11}}$}\right. \\ \tan \left(\raisebox{1ex}{$\mathrm{\theta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=\frac{3}{\sqrt{{{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2-6{\mathrm{x}}_1-4{\mathrm{y}}_1+4}} \\ \frac{3}{4}=\frac{3}{\sqrt{{{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2-6{\mathrm{x}}_1-4{\mathrm{y}}_1+4}} \\ {{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2-6{\mathrm{x}}_1-4{\mathrm{y}}_1-12=0 \\ By \mathrm{verifying} \mathrm{the} options ,(6,6) satisfies the above equation \end{gathered}$$