#### Answer

graph of the equation is nonexistent

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To determine whether the given equation, $
x^2+y^2+4x-8y+32=0
,$ is the graph of a circle, complete the square for both $x$ and $y$ variables and then analyze the value of the radius. If the value of the radius is positive, then the equation has a graph of a circle. If the radius is $0,$ then the equation describes a point. If the radius is negative, the graph is nonexistent.
$\bf{\text{Solution Details:}}$
Grouping the $x$ variables and the $y$ variables and also transposing the constant, the given equation is equivalent to
\begin{array}{l}\require{cancel}
(x^2+4x)+(y^2-8y)=-32
.\end{array}
With the coefficient of $x$ equal to $
4
,$ add $\left(\dfrac{
4
}{2}\right)^2=4
$ to both sides of the equation above to complete the square for $x$. That is,
\begin{array}{l}\require{cancel}
(x^2+4x+4)+(y^2-8y)=-32+4
\\\\
(x+2)^2+(y^2-8y)=-28
.\end{array}
With the coefficient of $y$ equal to $
-8
,$ add $\left(\dfrac{
-8
}{2}\right)^2=16
$ to both sides of the equation above to complete the square for $y$. That is,
\begin{array}{l}\require{cancel}
(x+2)^2+(y^2-8y+16)=-28+16
\\\\
(x+2)^2+(y-4)^2=-12
.\end{array}
Since $r^2=-12,$ is less than $0,$ the $\text{
graph of the equation is nonexistent
}.$