Let the equation of circle is as follows:-
$$
(x-a)^2+(y-b)^2=r^2
$$
Equation (i) is passing through the points $(2,0)$, $(1,-2) \&(-1,1)$. Then, we get
$$
\begin{aligned}
& (2-a)^2+(-b)^2=r^2 \\
& (1-a)^2+(-2-b)^2=r^2 \\
& (-1-a)^2+(1-b)^2=r^2
\end{aligned}
$$
Solving eqs. (ii), (iii) \& (iv), we get
$$
a=\frac{3}{14}, b=-\frac{5}{14} \& r^2=\frac{325}{98}
$$
Putting above values in eqn. (i), we get
$$
\begin{aligned}
& \left(x-\frac{3}{14}\right)^2+\left(y+\frac{5}{14}\right)^2=\frac{325}{98} \\
& \Rightarrow S \equiv\left(x-\frac{3}{14}\right)^2+\left(y+\frac{5}{14}\right)^2-\frac{325}{98}=0
\end{aligned}
$$
Now, $S(1,2)=\left(1-\frac{3}{14}\right)^2+\left(2+\frac{5}{14}\right)^2-\frac{325}{98}$
$$
\begin{aligned}
& =\frac{121}{196}+\frac{1089}{196}-\frac{325}{98} \\
& =\frac{121+1089-650}{196}=\frac{560}{196} \\
& \Rightarrow S(1,2)>0
\end{aligned}
$$
$\therefore(1,2)$ lies outside of the circle S.