## College Algebra (11th Edition)

$\text{Center: } (2,-3) \\\text{Radius: } 1$
$\bf{\text{Solution Outline:}}$ To find the Center-Radius Form of the given equation, $x^2+y^2-4x+6y+12=0 ,$ complete the square for both $x$ and $y$ variables. $\bf{\text{Solution Details:}}$ Grouping the $x$-variables and the $y$-variables and transposing the constant, the given equation is equivalent to \begin{array}{l}\require{cancel} (x^2-4x)+(y^2+6y)=-12 .\end{array} Since the coefficient of $x$ is $-4 ,$ add $\left(\dfrac{ -4 }{2}\right)^2= 4$ on both sides of the equation to complete the square for $x$. Hence, the equation above is equivalent to \begin{array}{l}\require{cancel} (x^2-4x+4)+(y^2+6y)=-12+4 \\\\ (x-2)^2+(y^2+6y)=-8 .\end{array} Since the coefficient of $y$ is $6 ,$ add $\left(\dfrac{ 6 }{2}\right)^2= 9$ on both sides of the equation to complete the square for $y$. Hence, the equation above is equivalent to \begin{array}{l}\require{cancel} (x-2)^2+(y^2+6y+9)=-8+9 \\\\ (x-2)^2+(y+3)^2=1 .\end{array} Using the Center-Radius Form of the equation of circles which is given by $(x-h)^2+(y-k)^2=r^2,$ where the center is $(h,k)$ and the radius is $r,$ the equation above is equivalent to \begin{array}{l}\require{cancel} (x-(2))^2+(y-(-3))^2=1^2 .\end{array} Hence the equation of the circle above has the following characteristics: \begin{array}{l}\require{cancel} \text{Center: } (2,-3) \\\text{Radius: } 1 .\end{array}