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

Equation of polar of point \(P(2,3)\) with respect to the circle \(x^2+y^2-2 x-2 y+1=0\) is \(\begin{aligned} & T=0 \Rightarrow 2 x+3 y-(x+2)-(y+3)+1=0 \\ & \Rightarrow \quad x+2 y-4=0 \quad \ldots (i) \end{aligned}\) End equation of line joining of points \(P(2,3)\) and centre of the given circle \(C(1,1)\) is \(y-1=\frac{2}{1}(x-1) \Rightarrow 2 x-y-1=0 \quad \ldots (ii)\) \(\because\) Inverse point of \(P\) with respect to the given circle is point of intersection of polar Eq. (i) and line CP, Eq. (ii), so coordinate of the inverse point is \(Q\left(\frac{6}{5}, \frac{7}{5}\right) \text {. }\) Now, equation of polar of point \(Q\left(\frac{6}{5}, \frac{7}{5}\right)\) with respect to given circle is \(\begin{aligned} & \frac{6}{5} x+\frac{7}{5} y-\left(x+\frac{6}{5}\right)-\left(y+\frac{7}{5}\right)+1=0 \\ \Rightarrow \quad & x+2 y-8=0 \quad \ldots (iii) \end{aligned}\) \(\therefore\) Required distance between polars (i) and (iii) is \(\frac{4}{\sqrt{1+4}}=\frac{4}{\sqrt{5}}\) Hence, option (b) is correct.