## University Calculus: Early Transcendentals (3rd Edition)

$(x - 0)^2+(y +1)^2+(z - 5)^2=4$
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 $x_0=0,y_0=-1,z_0=5, r=2$ Plug all the above values in equation (1), then we get Thus, $(x - 0)^2+(y -(-1))^2+(z - 5)^2=(2)^2$ Hence, $(x - 0)^2+(y +1)^2+(z - 5)^2=4$