Answer
Center: $(0,-\dfrac{1}{3},\dfrac{1}{3})$ and radius: $\dfrac{\sqrt {29}}{3}$
Work Step by Step
We know that the standard equation of a sphere is written as:
$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2\ \ \ $ (1)
Here, $(x_0,y_0,z_0)$ represents the center and $r$ is the radius of the sphere.
Since, we have $3x^2+3y^2+3z^2+2y-2z=9$
or, $2(x^2+\dfrac{1}{2}x)+2(y^2+\dfrac{1}{2}y)+2(z^2+\dfrac{1}{2}z)=9$
or, $3x^2+3(y+\dfrac{1}{3})^2+3(z-\dfrac{1}{3})^2=\dfrac{29}{3}$
or, $x^2+(y+\dfrac{1}{3})^2+(z-\dfrac{1}{3})^2=\dfrac{29}{9}$
or, $(x-0)^2+(y -(-\dfrac{1}{3}))^2+(z-\dfrac{1}{3})^2=(\sqrt{\dfrac{29}{9}})^2$
or, $(x-0)^2+(y -(-\dfrac{1}{3}))^2+(z-\dfrac{1}{3})^2=(\dfrac{\sqrt {29}}{3})^2$
Compare the above equation with equation (1), we get
Center :$(0,-\dfrac{1}{3},\dfrac{1}{3})$ and radius: $\dfrac{\sqrt {29}}{3}$
