Answer
Center: $(0,-\dfrac{1}{3},\dfrac{1}{3} )$ and radius: $\dfrac{4}{3}$
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 $x^2+(y + \dfrac{1}{3})^2+(z -\dfrac{1}{3})^2=\dfrac{16}{9}$
or, $(x - 0)^2+(y -(-\dfrac{1}{3}))^2+(z - \dfrac{1}{3})^2=(\sqrt {\dfrac{16}{9}})^2$
or, $(x - 0)^2+(y -(-\dfrac{1}{3}))^2+(z - \dfrac{1}{3})^2=(\dfrac{4}{3})^2$
Compare the above equation with equation (1):
Center: $(0,-\dfrac{1}{3},\dfrac{1}{3} )$ and radius: $\dfrac{4}{3}$
