The distance between the polar of P(2, 3) with respect to the circle $x^2+y^2-2x-2y+1=0$ and the polar of the inverse point of P with respect to the same circle is

$$\begin{gathered} \mathrm{S}={\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2\mathrm{y}+1=0 \\ centre \mathrm{C}=(1, 1),r=1 \\ Given point \mathrm{P}(2, 3) \end{gathered}$$

Equation of polar of P is ${\mathrm{S}}_1=0$

$$\begin{gathered} 2\mathrm{x}+3\mathrm{y}-(\mathrm{x}+2)-(\mathrm{y}+3)+1=0 \\ \mathrm{x}+2\mathrm{y}-4=0 -\left(1\right) \end{gathered}$$

Equation $\overline{\mathrm{CP}}$ is

$$\begin{gathered} \mathrm{y}-1=2 (\mathrm{x}-1) \\ \mathrm{y}-1=2\mathrm{x}-2 \\ 2\mathrm{x}-\mathrm{y}-1=0 -\left(2\right) \end{gathered}$$

Solving  (1) & (2) is

$$\begin{gathered} \mathrm{x}+2\mathrm{y}-4=0 \\ \underline{4\mathrm{x}-2\mathrm{y}-2=0} \\ 5\mathrm{x}-6=0 \\ \mathrm{x}=\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$5$}\right. , \mathrm{y}=\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$5$}\right. \end{gathered}$$

inverse point is Q $\left(\frac{6}{5}, \frac{7}{5}\right)$

 $$\begin{gathered} Polar of Q w.r.to S=0 is S_1=0 \\ \Rightarrow \mathrm{x}+2\mathrm{y}-8=0\to \left(3\right) \\ Now dis\tan ce between \left(1\right) and \left(3\right) is \\ =\frac{\left|-8+4\right|}{\sqrt{5}}=\frac{4}{\sqrt{5}} \end{gathered}$$