Answer
$\text{ graph of a circle with center }$ $
(2,-1)
\text{ and radius of }
3$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To evaluate whether the given equation, $
x^2+y^2-4x+2y=4
,$ has a circle as its graph, change the form of the equation to the Center-Radius Form. Then evaluate the value of $r^2.$ If the value of $r^2$ is positive, then the graph of the equation is a circle. If $r^2$ is equal to zero, then the equation describes a point. If $r^2$ is negative, the graph is nonexistent.
$\bf{\text{Solution Details:}}$
Grouping the $x$-variables and the $y$-variables, the given equation is equivalent to
\begin{array}{l}\require{cancel}
(x^2-4x)+(y^2+2y)=4
.\end{array}
Since the coefficient of $x$ is $
-4
,$ add $\left(\dfrac{
-4
}{2}\right)^2=
4
$ to both sides of the equation to complete the square for $x$. That is,
\begin{array}{l}\require{cancel}
(x^2-4x+4)+(y^2+2y)=4+4
\\\\
(x-2)^2+(y^2+2y)=8
.\end{array}
Since the coefficient of $y$ is $
2
,$ add $\left(\dfrac{
2
}{2}\right)^2=
1
$ to both sides of the equation to complete the square for $x$. That is,
\begin{array}{l}\require{cancel}
(x-2)^2+(y^2+2y+1)=8+1
\\\\
(x-2)^2+(y+1)^2=9
.\end{array}
In the form $(x-h)^2+(y-k)^2=r^2,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
(x-(2))^2+(y-(-1))^2=9
.\end{array}
Since $r^2=9,$ is greater than zero, then the equation has a $\text{ graph of a circle with center }$ $
(2,-1)
\text{ and radius of }
3
.$
