## Precalculus (10th Edition)

The center of the circle is $\left( -2,\,1 \right)$, and its radius is $3$. The graph of the circle is:
Consider ${{x}^{2}}+{{y}^{2}}+4x-2y-4=0$. By grouping the terms involving $x$ and involving $y$, and taking the constant to the right side, $\Rightarrow \left( {{x}^{2}}+4x \right)+\left( {{y}^{2}}-2y \right)=4$ Complete the square of each expression in parenthesis. For that, add ${{\left( \frac{4}{2} \right)}^{2}}={{\left( 2 \right)}^{2}}=4$ in parenthesis involving $x$ terms, and add ${{\left( \frac{2}{2} \right)}^{2}}={{\left( 1 \right)}^{2}}=1$ in parenthesis involving $y$ terms. Also add $4$ and $1$ to the left side. $\Rightarrow \left( {{x}^{2}}+4x+4 \right)+\left( {{y}^{2}}-2y+1 \right)=4+4+1$ $\Rightarrow \left( {{x}^{2}}+4x+4 \right)+\left( {{y}^{2}}-2y+1 \right)=9$ $\Rightarrow {{\left( x+2 \right)}^{2}}+{{\left( y-1 \right)}^{2}}={{3}^{2}}$ By comparing ${{\left( x+2 \right)}^{2}}+{{\left( y-1 \right)}^{2}}={{3}^{2}}$ with the standard form of the equation of the circle ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$, it gives $h=-2,\,k=1$ and $r=3$. Therefore, the center of the circle is $\left( -2,\,1 \right)$, and its radius is $3$. The four points on the graph are $\left( -2+3,\,1 \right)\equiv (1,\,1),\,\,\,\left( -2-3,\,1 \right)\equiv (-5,\,1),\,\,\left( -2,\,1+3 \right) \equiv (-2,\,4)$ $\left( -2,\,1-3 \right)\equiv (-2,\,-2)$. Thus the graph of the circle is shown above.