Answer
$$\eqalign{
& {\text{domain: }}\left[ {1,3} \right] \cr
& {\text{range: }}\left[ { - 4, - 2} \right] \cr} $$
Work Step by Step
$$\eqalign{
& {x^2} - 4x + {y^2} + 6y = - 12 \cr
& \left( {{x^2} - 4x + 4} \right) + \left( {{y^2} + 6y + 9} \right) = - 12 + 4 + 9 \cr
& {\text{Factoring}} \cr
& {\left( {x - 2} \right)^2} + {\left( {y + 3} \right)^2} = 1 \cr
& {\text{The equation is written in the form }}{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2} \cr
& {\left( {x - 2} \right)^2} + {\left( {y + 3} \right)^2} = 1,\,\,\,\,\,\,h = 2,\,\,\,\,k = - 3,\,\,\,\,\,r = 1 \cr
& {\text{This equation represents a circle centered at }}\left( {h,k} \right):\left( {2, - 3} \right) \cr
& {\text{Radius }}r = 1 \cr
& {\text{The domain of the circle is }}\left[ {h - r,h + r} \right] \cr
& {\text{domain: }}\left[ {1,3} \right] \cr
& {\text{The range of the circle is }}\left[ {k - r,k + r} \right] \cr
& {\text{range: }}\left[ { - 4, - 2} \right] \cr
& \cr
& {\text{Therefore,}} \cr
& {\text{domain: }}\left[ {1,3} \right] \cr
& {\text{range: }}\left[ { - 4, - 2} \right] \cr
& \cr
& {\text{Graph}} \cr} $$
