Answer
$(x + 1)^2+(y -\dfrac{1}{2})^2+(z+ \dfrac{2}{3})^2=\dfrac{16}{81}$
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 $x_0=-1,y_0=\dfrac{1}{2},z_0=\dfrac{-2}{3}, r=\dfrac{4}{9}$
Plug all the above values in equation (1), then we get
Thus,, $(x - (-1))^2+(y -(\dfrac{1}{2}))^2+(z - (\dfrac{-2}{3}))^2=(\dfrac{4}{9})^2$
Hence, $(x + 1)^2+(y -\dfrac{1}{2})^2+(z+ \dfrac{2}{3})^2=\dfrac{16}{81}$
