The equation of pole w.r.t. the point $(1,2)$ to the circle $x^2+y^2-4 x-6 y+9=0$ is
$$
\Rightarrow \quad \begin{aligned}
x+2 y-2(x+1)-3(y+2)+9 & =0 \\
\Rightarrow \quad x+y-1 & =0
\end{aligned}
$$
$\Rightarrow \quad x+y-1=0$ foot $(\alpha, \beta)$ of the perpendicular from the point $(1,2)$ to the line $x+y-1$.
$$
\begin{aligned}
& \therefore & \frac{\alpha-1}{1}=\frac{\beta-2}{1} & =-\frac{(1 \cdot 1+1 \cdot 2-1)}{1^2+1^2} \\
& \Rightarrow & \alpha-1 & =\beta-2=-1 \\
\Rightarrow & & \alpha & =0, \beta=1
\end{aligned}
$$
Hence, required point is $(0,1)$.