Answer
Standard Form: $x^2 + (y + \frac{1}{2})^2 = \frac{1}{4}$
General Fform: $x^2 + y^2 + y = 0.$
Refer to the graph below.
Work Step by Step
The standard form of a circle's equation is $(x−h)^2+(y−k)^2=r^2$ where $r$=radius and $(h,k)$ is the center.
Substitute $r=\frac{1}{2}, h=0, \text{ and } k=-\frac{1}{2}$ in the standard form above to obtain:
$$(x-0)^2+\left[y-\left(-\frac{1}{2}\right)\right]^2=\left(\frac{1}{2}\right)^2\\
x^2+\left(y+\frac{1}{2}\right)^2=\frac{1}{4}$$
The general form of the equation of a circle is $x^2+y^2+ax+by+c=0$..
Subtract $\frac{1}{4}$ from each side of the equation above to obtain:
\begin{align*}
x^2+\left(y+\frac{1}{2}\right)^2−\frac{1}{4}=0\\
x^2+y^2+y+\frac{1}{4}−\frac{1}{4}=0\\
x^2+y^2+y=0
\end{align*}
With $r=\frac{1}{2}$ and center at $(0, -\frac{1}{2})$, plot the points $\frac{1}{4}$ unit directly above, below, to the left, and to the right of the circle$. Then, connect these four points using a smooth curve to form a circle.
Refer to the graph above.
