#### Answer

$\text{Center: }
(2,-3)
\\\text{Radius: }
1$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
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}