Answer
does not have a circle as its graph
Work Step by Step
$\bf{\text{Solution Outline:}}$
To evaluate whether the given equation, $
x^2+y^2+6x+10y+36=0
,$ 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 and transposing the constant, the given equation is equivalent to
\begin{array}{l}\require{cancel}
(x^2+6x)+(y^2+10y)=-36
.\end{array}
Since the coefficient of $x$ is $
6
,$ add $\left(\dfrac{
6
}{2}\right)^2=
9
$ to both sides of the equation to complete the square for $x$. That is,
\begin{array}{l}\require{cancel}
(x^2+6x+9)+(y^2+10y)=-36+9
\\\\
(x+3)^2+(y^2+10y)=-27
.\end{array}
Since the coefficient of $y$ is $
10
,$ add $\left(\dfrac{
10
}{2}\right)^2=
25
$ to both sides of the equation to complete the square for $x$. That is,
\begin{array}{l}\require{cancel}
(x+3)^2+(y^2+10y+25)=-27+25
\\\\
(x+3)^2+(y+5)^2=-2
.\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-(-3))^2+(y-(-5))^2=-2
.\end{array}
Since $r^2=-2,$ is less than zero, then the equation $\text{
does not have a circle as its graph
}.$
