If the complex number z=x+iy satisfies the condition |z+1|=1 , then z lies on
∵|z+1|=1
⇒|x+iy+1|=1
⇒|x+1+iy|=1
⇒√(x+1)2+y2=1
⇒(x+1)2+y2=12
(squaring both sides)
Clearly, above equation represents the equation of a circle.
Centre (−1,0) and radius =1
Hence, option (B) is correct.