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

$\begin{array}{rl} & s=x^2+y^2-4x-6y+9=0\\ & c=(2,3)P(1,2)\end{array}$
Q is inverse point of “P”.
Polar of P w.r.t. to  is $s_1=0$
$x+y-1=0$
Q is foot of the perpendicular of P on the  
line $x+y-1=0$
$\begin{array}{rl} & Q=(0,1)\\ & PQ=\sqrt{2}\end{array}$