$$
\begin{aligned}
& \text { } \because|z|=1,|z-2|=1,|z-1|=0 \\
& \because \quad|z|=1 \Rightarrow x^2+y^2=1...(i) \\
& |z-2|=1 \Rightarrow(x-2)^2+(y)^2=1 ...(ii) \\
& |z-1|=0 \Rightarrow(x-1)^2+y^2=0...(iii)
\end{aligned}
$$
Solving Eqs. (i) and (ii), we get
$$
\begin{aligned}
\quad(x-2)^2-x^2 & =0 \\
\Rightarrow \quad(x-2)^2 & =x^2 \Rightarrow x-2= \pm x
\end{aligned}
$$
Taking positive
$$
\begin{aligned}
x-2 & =x \\
-2 & =0 \text { (Not true) }
\end{aligned}
$$
Taking negative
$$
\begin{aligned}
x-2 & =-x \\
2 x & =2 \\
x & =1 \Rightarrow y^2=1-x^2=1-1=0 \\
y & =0
\end{aligned}
$$
$\therefore$ Point of intersection of Eqs. (i) and (ii) is $(1,0)$.
Now, put it in Eq. (iii), we get
$$
\begin{aligned}
(1-1)^2+0^2 & =0 \\
0 & =0
\end{aligned}
$$
$\therefore$ It also lies on circle of Eq. (iii).
$\therefore$ Common point is $(1,0)$.