Let the circle
$\begin{aligned}
& \mathrm{S} \equiv x^2+y^2+2 g x+2 f y+c=0 \\
& x^2+y^2-4 x-2 y+4=0 \\
& x^2+y^2-2 x-4 y+1=0
\end{aligned}$
and $x^2+y^2+4 x+2 y+1=0$, respectively.
$\begin{aligned}
& \therefore \quad 2 g(2)+2 f(1)=c+4 \\
& \Rightarrow \quad 4 g+2 f=c+4... (i) \\
& 2 g(1)+2 f(2)=c+1 \\
& \Rightarrow \quad 2 g+4 f=c+1... (ii)
\end{aligned}$
and $2 g(-2)+2 f(-1)=1+c$
$\Rightarrow-4 g-2 f=1+c... (iii)$
On solving eqs. (i), (ii) and (iii), we get
$g=\frac{3}{4}, f=-\frac{3}{4}, c=-\frac{10}{4}$
$\therefore \quad$ Equation of circle,
$S=x^2+y^2+\frac{3}{2} x-\frac{3}{2} y-\frac{10}{4}=0$
$\begin{aligned} & \therefore \text { Radius }=\sqrt{\left(\frac{3}{4}\right)^2+\left(\frac{-3}{4}\right)^2+\frac{10}{4}} \\ & =\sqrt{\frac{9}{16}+\frac{9}{16}+\frac{10}{4}}=\sqrt{\frac{58}{16}}=\sqrt{\frac{29}{8}}\end{aligned}$