The correct options are
A (1,2√2)
B (1+2√2,0)
C (1,−2√2)
D (1−2√2,0)
x2−y2−2x+1=0⇒(x−1)2−y2=0⇒(x−1+y)(x−1−y)=0⇒x+y=1; x−y=1
Hence centre of the required circle will lie on the angle bisector of above lines i.e., y=0; x=1
Case 1: When centre lies on x=1
Let centre be (1,a)
Then its distance from x+y=1 will be radius i.e., 2 units
∴|1+a−1|√2=2⇒a=±2√2
Hence centres will be (1,±2√2)
Case 2: When centre lies on y=0
Let centre be (b,0)
Similar with case 1
∴|b−1|√2=2⇒b=1±2√2
Hence centres will be (1±2√2,0)