## College Algebra (11th Edition)

$\bf{\text{Solution Outline:}}$ 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 $\text{ does not have a circle as its graph }.$