If the inverse of P(1, 2) w.r.t to $x^2+y^2-4x-6y+9=0$ is Q then PQ = 
 

$$\begin{gathered} Given circle S\equiv x^2+y^2-4x-6y+9=0 \\ For given circle \\ centre\left(C\right)=(2,3) \\ Given point P=(1,2) \\ Now equation of line CP is x-y+1=0\to \left(1\right) \\ Now polar of P is x+y-1=0\to \left(2\right) \\ Now point of intersection of (1 )and \left(2\right) is Q=(0,1) \\ Now PQ=\sqrt{2} \end{gathered}$$