We have General Equation of Circle is given as

x^{2}+y^{2}+2gx+2fy+c=0 ---(1)

Since, Circle passes through, (0,0) and (1,0)

Therefore Equation (1) reduces to,

c=0 ---(2)

1+2g+c=0 ---(3),

From (2) and (3), g= -1/2

Also, From As per Question,

Cicle in equation, (1) touches, another circle having equation,

x^{2}+y^{2}=(3)^{2} ---(4)

SInce the center (0,0) of circle (4) lies on the loci of the circle (1), therefore, The radius of the circle (4) must be equal to diameter of circle(1)

hence, 2√g^{2}+f^{2}-c= 3

=> √1/4+f^{2}=3/2

=> f^{2}= 9/4-1/4

=> f^{2}= 2

=> f= ±√2

**Hence Centres of Circle (1) is (-g,-f) or (1/2,±√2)**

**Option (B) is correct**