## Precalculus (10th Edition)

Standard Form: $(x - \frac{1}{2}) ^2 + y^2 =\frac{1}{4}$ General Form: $x^2 +y^2 - x = 0$ 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 $r=\frac{1}{2}, h=\frac{1}{2}, \text{ and } k=0$ in the standard form above to obtain: $$\left(x−\frac{1}{2}\right)^2+(y-0)^2=\left(\frac{1}{2}\right)^2\\ \left(x-\frac{1}{2}\right)^2+y^2=\frac{1}{4}$$ 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*} \left(x-\frac{1}{2}\right)^2+y^2−\frac{1}{4}&=0\\ x^2-x+\frac{1}{4}+y^2-\frac{1}{4}&=0\\ x^2+y^2-x&=0 \end{align*} With $r=\frac{1}{2}$ and center at $\left(\frac{1}{2},0\right)$, plot the points $\frac{1}{2}$ units 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.