Answer
$${\text{Sphere centered at }}\left( { - 1,1,0} \right){\text{ with radius }}5$$
Work Step by Step
$$\eqalign{
& {\left( {x + 1} \right)^2} + {y^2} + {z^2} - 2y - 24 = 0 \cr
& {\text{Group terms}} \cr
& {\left( {x + 1} \right)^2} + \left( {{y^2} - 2y} \right) + {z^2} = 24 \cr
& {\text{Complete the square}} \cr
& {\left( {x + 1} \right)^2} + \left( {{y^2} - 2y + 1} \right) + {z^2} = 24 + 1 \cr
& {\left( {x + 1} \right)^2} + {\left( {y - 1} \right)^2} + {z^2} = {\left( 5 \right)^2} \cr
& {\text{The equation is in the form }}{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} + {\left( {z - c} \right)^2} = {r^2} \cr
& {\text{Represents a sphere centered at }}\left( {a,b,c} \right){\text{ with radius }}r,{\text{ then}} \cr
& {\left( {x + 1} \right)^2} + {\left( {y - 1} \right)^2} + {z^2} = {\left( 5 \right)^2} \cr
& {\text{Sphere centered at }}\left( { - 1,1,0} \right){\text{ with radius }}5 \cr} $$
