∵|z|=1,|z−2|=1,|z−1|=0 ∵|z|=1 ⇒x2+y2=1...(i) |z−2|=1 ⇒(x−2)2+(y)2=1 ...(ii) |z−1|=0 ⇒(x−1)2+y2=0 ...(iii) Solving Eqs. (i) and (ii), we get (x−2)2−x2=0 Taking positive x−2=x −2=0 (Not true) Taking negative x−2=−x 2x=2 x=1 ⇒y2=1−x2=1−1=0 y=0 ∴ Point of intersection of Eqs. (i) and (ii) is (1,0). Now, put it in Eq. (iii), we get (1−1)2+02=0 0=0 ∴ It also lies on circle of Eq. (iii). ∴ Common point is (1,0).