The equation of the circle with centre (3, 2) and the power of (1, –2) w.r.t the circle $x^2+y^2=1$  as radius is

Given that centre $(3,2)=(h,k)$ and equation circle $x^2+y^2=1$…… (1)

and point P(1,-2)

Since radius (r)= power of P(1,-2) with respect to (1)

$\begin{array}{c}=S_{11}\\ =(1{)}^2+(-2{)}^2-1\\ r=4\end{array}$

 Equation of the circle having centre (h,k) and radius r is

$\begin{array}{l}(x-h{)}^2+(y-k{)}^2=r^2\\ \Rightarrow (x-3{)}^2+(y-2{)}^2=(4{)}^2\\ \Rightarrow x^2+y^2-6x-4y-3=0\end{array}$

 Equation of the circle is $x^2+y^2-6x-4y-3=0$