If the complex number z=x+iy satisfies the condition

$|z+1|=1$

$\Rightarrow|z+1|^{2}=1^{2}$

$\Rightarrow(z+1) \overline{(z+1)}=1$

$\Rightarrow(z+1)(\bar{z}+1)=1$

$\Rightarrow z \bar{z}+z+\bar{z}+1=1$

$\Rightarrow z \bar{z}+z+\bar{z}=0$

Since, $z=x+i y$

$\therefore z \bar{z}+z+\bar{z}=0$

$\Rightarrow(x+i y)(x-i y)+x+i y+x-i y=0$

$\Rightarrow x^{2}+y^{2}+2 x=0$

$\Rightarrow(x+1)^{2}+(y-0)^{2}=1^{2}$

which is the equation of a circle with centre $(-1,0)$ and radius 1

Hence, the correct option is (b)