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(B) The equation of the polar of point $P(2,3)$ with respect to the circle $x^2+y^2-2x-2y+1=0$ is given by $T=0$: $x(2)+y(3)-(x+2)-(y+3)+1=0$ $x+2y-4=

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$\frac{4}{\sqrt{5}}$

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(B) The equation of the polar of point $P(2,3)$ with respect to the circle $x^2+y^2-2x-2y+1=0$ is given by $T=0$: $x(2)+y(3)-(x+2)-(y+3)+1=0$ $x+2y-4=0$ $\ldots(i)$ The center of the circle is $C(1,1)$ and the radius is $r = \sqrt{1^2+1^2-1} = 1$. The line joining $P(2,3)$ and $C(1,1)$ is $y-1 = \frac{3-1}{2-1}(x-1)$,which simplifies to $y-1 = 2(x-1)$ or $2x-y-1=0$ $\ldots(ii)$. The inverse point $Q$ of $P$ lies on the line $CP$ and the polar of $P$. Solving $(i)$ and $(ii)$: $x+2(2x-1)-4=0$ $\Rightarrow 5x-6=0$ $\Rightarrow x = \frac{6}{5}$. $y = 2(\frac{6}{5})-1 = \frac{7}{5}$. So $Q = (\frac{6}{5}, \frac{7}{5})$. The polar of $Q(\frac{6}{5}, \frac{7}{5})$ is: $\frac{6}{5}x + \frac{7}{5}y - (x+\frac{6}{5}) - (y+\frac{7}{5}) + 1 = 0$ $\frac{1}{5}x + \frac{2}{5}y - \frac{8}{5} = 0 \Rightarrow x+2y-8=0$ $\ldots(iii)$. The distance between parallel lines $x+2y-4=0$ and $x+2y-8=0$ is: $d = \frac{|-4 - (-8)|}{\sqrt{1^2+2^2}} = \frac{4}{\sqrt{5}}$. Thus,option $(b)$ is correct.

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

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