Answer
$\text{ graph of a circle with center }$ $
(-1,-8)
\text{ and radius of }
2$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To evaluate whether the given equation, $
x^2+y^2+2x+16y=-61
,$ 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+2x)+(y^2+16y)=-61
.\end{array}
Since the coefficient of $x$ 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+2x+1)+(y^2+16y)=-61+1
\\\\
(x+1)^2+(y^2+16y)=-60
.\end{array}
Since the coefficient of $y$ is $
16
,$ add $\left(\dfrac{
16
}{2}\right)^2=
64
$ to both sides of the equation to complete the square for $x$. That is,
\begin{array}{l}\require{cancel}
(x+1)^2+(y^2+16y+64)=-60+64
\\\\
(x+1)^2+(y+8)^2=4
.\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-(-1))^2+(y-(-8))^2=2^2
.\end{array}
Since $r=2,$ is greater than zero, then the equation has a $\text{ graph of a circle with center }$ $
(-1,-8)
\text{ and radius of }
2
.$
