## Precalculus (10th Edition)

Standard Form $(x - 2)^2 + (y + 3)^2 = 16$ General Form: $x^2 + y^2 - 4x + 6y - 3 = 0$ 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 $r=4$, $h=2$, and $k=-3$in the standard form above to obtain: \begin{align*} (x-2)^2+[y-(-3)]^2&=4^2\\ (x-2)^2+(y+3)^2&=16 \end{align*} General form of the equation of a circle is $x^2+y^2 + ax+by+c=0$. Subtract $16$ from each side of the equation above to obtain: \begin{align*} (x-2)^2+(y+3)^2-16&=16-16\\ x^2-4x+4+y^2+6y+9-16&=0\\ x^2+y^2-4x+6y-3&=0 \end{align*} With $r=4$ and center at $(2, -3)$, plot the points $4$ 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.