Answer
The equation represents a circle
$ (x+\frac{1}{4})^2+(y^2+1)^2=1^2$
with center $(-\frac{1}{4},-1)$ and radius is $1$
Work Step by Step
$\underline{\textbf{Given equation}}$
$x^2+y^2+\frac{1}{2}x+2y+\frac{1}{16}=0$
$\Rightarrow x^2+\frac{1}{2}x+y^2+2y=-\frac{1}{16}$
$\Rightarrow (x^2+\frac{1}{2}x+\frac{1}{16})+(y^2+2y+1)=-\frac{1}{16}+1+\frac{1}{16}$
$\Rightarrow (x^2+\frac{1}{2}x+\frac{1}{16})+(y^2+2y+1)=1$
$\Rightarrow (x+\frac{1}{4})^2+(y^2+1)^2=1^2$
Which is the standard form for the equation of circle with center $(-\frac{1}{4},-1)$ and radius is $1$
