### Video Transcript

A sphere of radius 50 is centered on the point on the π§-axis that is at a distance of 17 from the π₯π¦-plane. What is the equation of the sphere?

In this question, weβre asked to determine the equation of a sphere. Weβre told that the radius of the sphere is equal to 50. And weβre also given some information about the center of the sphere. The center of the sphere is the point on the π§-axis which is a distance of 17 from the π₯π¦-plane. So to answer this question, letβs start by recalling how we represent a sphere with an equation.

We recall a sphere centered at the point π, π, π with a radius of π will have the equation π₯ minus π all squared plus π¦ minus π all squared plus π§ minus π all squared is equal to π squared. We call this the standard form for the equation of a sphere. To find this equation of our sphere, all we need to know is the center of our sphere. And we also need to know the radius of the sphere.

In the question, weβre already told that the radius of the sphere is equal to 50. So we can substitute π is equal to 50 into this equation. Then all we need to find is the coordinates of the center of the sphere. To do this, weβre going to need to use the fact that the center of the sphere is the point on the π§-axis which is a distance of 17 from the π₯π¦-plane.

To help us find this point, letβs sketch the information weβre given in the diagram. Weβll start by sketching the π₯π¦-plane on a set of three-dimensional axes, where weβll have our π§-axis in line with the screen. We need to find the points on the π§-axis which are a distance of 17 from this plane. Remember, when weβre talking about the distance between a point and a plane, we actually mean the perpendicular distance between the point and the plane.

This is actually really useful in this case because the π₯π¦-plane must be perpendicular to the π§-axis. Therefore, if we want to find the perpendicular distance between a point on the π§-axis and the π₯π¦-plane, we would just need to find its π§-coordinate because the π§-axis is already perpendicular with the π₯π¦-plane. Therefore, the point on the π§-axis with π§-coordinate 17 will be a distance of 17 from the π₯π¦-plane.

However, this is not the only option. We could also have the point on the π§-axis with π§-coordinate negative 17. The perpendicular distance between this point and the π₯π¦-plane is also 17. This gives us two possible points for the center of our sphere, the point zero, zero, 17 or the point zero, zero, negative 17. Both of these will give us valid equations for the sphere based on the information given to us in the question. So weβll need to substitute both of these into the equation of our sphere.

Substituting the point zero, zero, 17 as the center of our sphere and a radius of 50 into the standard form of the equation of a sphere, we get the equation π₯ minus zero all squared plus π¦ minus zero all squared plus π§ minus 17 all squared is equal to 50 squared. And if we evaluate and simplify this expression, we get π₯ squared plus π¦ squared plus π§ minus 17 all squared is equal to 2,500.

We can do exactly the same with the point zero, zero, negative 17 as the center and the radius of 50. This gives us an alternate equation of π₯ squared plus π¦ squared plus π§ plus 17 all squared is equal to 2,500. Both of these will be valid equations of a sphere of radius 50 centered at a point on the π§-axis which is at a distance of 17 from the π₯π¦-plane.