## Precalculus (10th Edition)

Standard Form: $(x + 2)^2 + (y - 1)^2 = 16$ General Form: $x^2 + y^2 + 4x - 2y - 11 = 0$. 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 $r=4, h=-2, \text{ and } k=1$ in the standard form above to obtain: $$[x−(-2)]^2+(y-1)^2=4^2\\ (x+2)^2+(y-1)^2=16$$ The 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-1)^2−16=0\\ x^2+4x+4+y^2-2y+1−16=0\\ x^2+y^2+4x-2y−11=0 \end{align*} With $r=4$ and center at $(-2,1)$, 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.