## Precalculus (10th Edition)

Standard Form: $x^2+y^2=9$ General Form. $x^2+y^2-9=0$ Refer to the graph below.
The standard form of acircle's equation is $(x-h)^2+(y-k)^2=r^2$ where $r$=radius and $(h, k)$ is the center. Substitute $3$ to $r$ and $0$ to both $h$ and $k$ in the standard form above to obtain: \begin{align*} (x-0)^2+(y-0)^2&=3^2\\ x^2+y^2&=9 \end{align*} General form of the equation of a circle is $x^2+y^2 + ax+by+c=0$. Subtract $9$ from each side of the equation above to obtain: \begin{align*} x^2+y^2-9&=9-9\\ x^2+y^2-9&=0 \end{align*} With $r=3$ and center at $(0, 0)$, plot the points $3$ 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.