Answer
$\text{ graph of a circle with center }$ $
(6,0)
\text{ and radius of }
4$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To evaluate whether the given equation, $
x^2+y^2-12x+20=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^2-12x)+y^2=-20
.\end{array}
Since the coefficient of $x$ is $
-12
,$ add $\left(\dfrac{
-12
}{2}\right)^2=
36
$ to both sides of the equation to complete the square for $x$. That is,
\begin{array}{l}\require{cancel}
(x^2-12x+36)+y^2=-20+36
\\\\
(x-6)^2+y^2=16
.\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-(6))^2+(y-0)^2=4^2
.\end{array}
Since $r^2=16,$ is greater than zero, then the equation has a $\text{ graph of a circle with center }$ $
(6,0)
\text{ and radius of }
4
.$
