## Precalculus (10th Edition)

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.
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.