### Video Transcript

Find the equation of the sphere with centre two, negative three, four that touches the π₯π¦-plane.

Now, we know the general equation of a sphere the sphere with centre π, π, π and radius π has equation π₯ minus π squared plus π¦ minus π squared plus π§ minus π squared equals π squared. And we know that the coordinates of the centre of our sphere are two, negative three, four. So we can substitute these values for π, π, and π in our general equation. To find that, the equation of our sphere must have the form π₯ minus two squared plus π₯ minus negative three squared plus π§ minus four squared equals π squared.

The only thing we have left to find is the radius π. Weβre not given the value of the radius of the sphere explicitly in the question. But we are given some information that we havenβt used yet, that the sphere touches the π₯π¦-plane. Perhaps, we can use this fact to find the radius of our sphere. Letβs use the diagram to help us see whatβs going on.

Weβre working in three dimensions. And so, there are three axes: the π₯-, π¦-, and π§-axes. But itβs a bit hard to draw accurately in three dimensions. So what we can do is take a side-on view, where the π¦-axis is pointing at directly into the screen away from you. And so now whatβs looks like just the π₯-axis is actually the π₯π¦-plane seen side-on. Now, what does our sphere look like in this setting? Its centre is two, negative three, four. But as the π¦-axis is disappearing into the screen, only the π₯- and π§-coordinates are important for our diagram. The π₯-coordinate of our centre is two and the π§-coordinate is four. So our centre lies here.

In this view, a sphere with centre at this point will just look like a circle. Now, clearly, this sphere is too small it doesnβt get anywhere near the π₯π¦-plane and it is supposed to touch it. On the other hand, this sphere is too large. It intersects our plane in what looks like two points on the diagram. But in reality, remembering the hidden π¦-direction, we actually get an entire circle of intersection. Weβd like our sphere just to touch the π₯π¦-plane in one point. This is the sphere that we want. It touches the π₯π¦-plane only intersecting in a single point.

Now, we can find its radius. We can see that the radius of our sphere is four. The π§-coordinate of the centre of the sphere is four and the π§-coordinate of the point directly below it where the sphere touches the π₯π¦-plane is zero. And four minus zero is four. And itβs not surprising that the radius just turned out to be the π§-coordinate of our centre.

For a general point π, π, π in 3D space, the π₯-coordinate π is the distance to this point from the π¦π§-plane, the π¦-coordinate π is the distance to this point from the π₯π§-plane, and the π§-coordinate π is the distance of the point from the π₯π¦-plane. This is like how the π₯-coordinate of a point in 2D space represents its distance from the π¦-axis and the π¦ coordinate of the point in 2D space represents its distance from the π₯-axis.

Going back to three dimensions then, we see that the distance of our centre with coordinate two, negative three, four from the π₯π¦-plane is just its π§-coordinate four. And so, naturally, this will be the radius of the sphere with centre two, negative three, four that touches the π₯π¦-plane. Anyway, have we found that π is four, we need to substitute it into our equation. π₯ minus two squared plus π¦ minus negative three squared plus π§ minus four squared is then equal to four squared. On the left-hand side, we can simplify a bit writing π¦ minus negative three as π¦ plus three. And on the right-hand side, four squared is 16.

And this is our answer: π₯ minus two squared plus π¦ plus three squared plus π§ minus four squared is 16.