Lesson Video: Equation of a Sphere Mathematics

In this video, we will learn how to find the equation of a sphere given its center and how to find the center and the radius given the sphere’s equation.

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Video Transcript

Equation of a Sphere

In this lesson, we’re going to learn the standard form for the equation of a sphere. And we’re going to learn how to find the standard form for the equation of a sphere given the center of our sphere and the radius of our sphere. And we’ll also see how to find the center and radius of our sphere given the equation of the sphere in standard form.

So to find the equation of a sphere, we’re first going to need to recall exactly what we mean by a sphere. We recall a sphere centered at the point 𝑃 with a radius of 𝑟 will consist of all points a distance of 𝑟 units away from the point 𝑃. And there’s a lot of things we’re pointing out about this definition. First, spheres are in three dimensions. So, in particular, all of the points on our sphere will be given in three-dimensional coordinates and the point 𝑃 will also be given in three-dimensional coordinates. And it’s important we remember this because if we instead use two dimensions, this would just be the definition of a circle.

Next, since the radius 𝑟 represents a length or a distance, we say that 𝑟 must be positive. If our radius 𝑟 was equal to zero, then our sphere would just be all points at a distance of zero from 𝑃. It would just be the point 𝑃 itself. And if 𝑟 was negative, our definition wouldn’t make sense. So we just said 𝑟 has to be positive. So now that we have the definition of our sphere, we want to find the equation of our sphere. And there’s a few steps to doing this. Remember, for the equation for our sphere to be correct, every point on our sphere must satisfy our equation, and every point which satisfies our equation must lie on the sphere.

And we want to find the equation for a general sphere. So we’ll set the center of our sphere to be the point 𝑎, 𝑏, 𝑐 and we’ll let 𝑄, the point 𝑥, 𝑦, 𝑧, lie on our sphere of radius 𝑟 centered at the point 𝑃. And this is the general sphere we’re going to try and find the equation of. One way of finding the equation of this sphere would be to copy the method we use to find the equation of a circle. We drew a picture of our circle and then used what we knew about the Pythagorean theorem to find an equation for our circle. In fact, this method would work in exactly the same way. It would be slightly more complicated because we would be working in three dimensions. However, this would give us the correct equation for a sphere. However, we actually have more powerful tools to deal with this, giving us an easier method to find the equation of the sphere.

Remember, every point on our sphere must be a distance of 𝑟 from our point 𝑃. In our case, the point 𝑄 must be a distance 𝑟 from the center of our sphere 𝑃. And we know a formula to calculate the distance between two points in three dimensions. We recall, to calculate the distance between the point 𝑥 one, 𝑦 one, 𝑧 one and the point 𝑥 two, 𝑦 two, 𝑧 two, we just calculate the square root of 𝑥 two minus 𝑥 one all squared plus 𝑦 two minus 𝑦 one all squared plus 𝑧 two minus 𝑧 one all squared. So let’s use this formula to calculate the distance between point 𝑃 and the point 𝑄. Remember, we already know that this distance is going to be equal to 𝑟. This gives us the distance between the point 𝑃 and the point 𝑄 is the square root of 𝑥 minus 𝑎 all squared plus 𝑦 minus 𝑏 all squared plus 𝑧 minus 𝑐 all squared. And we know this is going to be equal to the radius of our sphere 𝑟.

So let’s think about exactly what we’ve just shown. We’ve shown any point 𝑄 which lies on the sphere of radius 𝑟 centered at our point 𝑃 must satisfy this equation. And in fact, we can show the opposite is true as well. We can think about what would happen if the point 𝑥 zero, 𝑦 zero, 𝑧 zero satisfied this equation. This means we substitute 𝑥 equals 𝑥 zero, 𝑦 equals 𝑦 zero, and 𝑧 equals 𝑧 zero into this equation. And the left-hand side of this equation will end up being equal to 𝑟. However, we can also see on the left-hand side of this equation, we’re just calculating the distance between the point 𝑥 zero, 𝑦 zero, 𝑧 zero and the center of our sphere 𝑃. Therefore, if the point satisfies this equation, it must be a distance 𝑟 from the center of our sphere 𝑃, which means it lies on our sphere.

Therefore, we’ve managed to find the equation of a sphere centered at the point 𝑎, 𝑏, 𝑐 with a radius of 𝑟. We’ve shown every point on this sphere must satisfy this equation and every point which satisfies this equation must lie on our sphere. However, there’s one thing you might have noticed. This is a very complicated-looking expression. In fact, we can simplify this by squaring both sides. And there is one thing worth pointing out here. We can justify that we won’t be gaining extra solutions in this case because we know our value of 𝑟 is positive. And everything inside of our square root symbol on the left-hand side of the equation is greater than or equal to zero because we’re taking the square of each term. This means we can guarantee squaring both sides will not give us extra solutions.

And this gives us the following equation for our sphere. We call this the standard form for the equation of a sphere, just like we might call this equation the standard form for the equation of a circle if we did not have the third term on the left-hand side. So let’s clear some space and think about exactly what we’ve just shown. We have shown the standard form for the equation of a sphere centered at the point 𝑎, 𝑏, 𝑐 with a radius of 𝑟 is given by 𝑥 minus 𝑎 all squared plus 𝑦 minus 𝑏 all squared plus 𝑧 minus 𝑐 all squared is equal to 𝑟 squared. So now, given the center of our sphere and the radius of our sphere, we can find the equation of our sphere. And equally, if we were given the standard form for the equation of our sphere, we would be able to find the center of the sphere and the radius of our sphere. Let’s now see a few examples of how we can apply this.

Give the equation of the sphere of center 11, eight, negative five and radius three in standard form.

In this question, we’re asked to find the equation of a sphere. And not only this, we’re asked to find this equation in standard form. We’re also given some information about our sphere. We’re told the center of our sphere is 11, eight, negative five and the radius of our sphere is three. To answer this question, let’s start by recalling what we mean by the standard form of the equation of a sphere. We recall a sphere centered at the point 𝑎, 𝑏, 𝑐 with a radius of 𝑟 will have the following equation in standard form: 𝑥 minus 𝑎 all squared plus 𝑦 minus 𝑏 all squared plus 𝑧 minus 𝑐 all squared is equal to 𝑟 squared.

In the question, we’re told the center of our sphere is the point 11, eight, negative five, and we’re told the radius of our sphere is three. So all we need to do to answer this question is substitute these values into the standard form for the equation of a sphere. Substituting 𝑎 is equal to 11, 𝑏 is equal to eight, 𝑐 is equal to negative five, and 𝑟 is equal to three into our equation for a sphere, we get 𝑥 minus 11 all squared plus 𝑦 minus eight all squared plus 𝑧 minus negative five all square is equal to three squared.

And of course we can simplify this. 𝑧 minus negative five all squared is equal to 𝑧 plus five all squared, and three squared is equal to nine. And this gives us our final answer. The standard form for the equation of a sphere centered at the point 11, eight, negative five with a radius of three is given by 𝑥 minus 11 all squared plus 𝑦 minus eight all squared plus 𝑧 plus five all squared is equal to nine.

Let’s now see an example where we’re given the equation of a sphere and we need to find the sphere’s center and radius.

Given that a sphere’s equation is 𝑥 plus five all squared plus 𝑦 minus 12 all squared plus 𝑧 minus two all squared minus 289 is equal to zero, determine its center and radius.

In this question, we’re given an equation, and we’re told that this equation represents a sphere. We need to determine the center of this sphere and the radius of this sphere. To do this, we can look at the equation we’re given, and we can see it’s very similar to the standard form for the equation of a sphere. So let’s start by recalling the standard form for the equation of a sphere. We recall a sphere of radius 𝑟 centered at the point 𝑎, 𝑏, 𝑐 will have the following equation in standard form. 𝑥 minus 𝑎 all squared plus 𝑦 minus 𝑏 all squared plus 𝑧 minus 𝑐 all squared is equal to 𝑟 squared. Therefore, if we can rewrite the equation we’re given in the question into standard form, we’ll be able to find the center of our sphere and we’ll also be able to find the radius of our sphere.

So let’s start with the equation of the sphere given to us in the question. We can see in the standard form for the equation of our sphere, our constant is on the right-hand side of our equation. However, in our equation, we can see it’s on the left. So the first thing we’re going to need to do is add 289 to both sides of our equation. This gives us 𝑥 plus five all squared plus 𝑦 minus 12 all squared plus 𝑧 minus two all squared is equal to 289. And now we can see our equation is almost in the correct form. However, because we want to find the radius of this sphere, we’re going to write our constant on the right-hand side as a square. And by taking the square root of 289, we can see that 289 is equal to 17 squared. So we can just rewrite the right-hand side of this equation as 17 squared.

And now our equation is almost exactly in standard form. However, we can see in the standard form, we need to subtract the constants inside of our parentheses. However, on the left-hand side of our equation, inside our parentheses, we have 𝑥 plus five. And we can fix this by realizing 𝑥 plus five is the same as 𝑥 minus negative five. So we’ll rewrite this term as 𝑥 minus negative five all squared. And now that we’ve written this equation in the standard form for the equation of a sphere, we can find the center and radius of this sphere. The center of our sphere will be the point negative five, 12, two because these are the values of 𝑎, 𝑏, and 𝑐 in our standard form for the equation of a sphere. And another way of thinking about this is these are the values of 𝑥, 𝑦, and 𝑧 which will make each term on the left-hand side of our equation equal to zero. And we can also find the radius of this sphere. The radius of this sphere is going to be equal to 17.

And remember, the radius of a sphere represents a length, so we can give this units. We’ll call this 17 length units. Therefore, given the equation of the sphere is 𝑥 plus five all squared plus 𝑦 minus 12 all squared plus 𝑧 minus two all squared minus 289 is equal to zero, by rewriting this equation into the standard form for the equation of a sphere, we were able to find its center and radius. We were able to show the center of the sphere was the point negative five, 12, two and the radius of this sphere was 17 length units.

Let’s now see an example where we check whether a given equation is the equation of a sphere.

Determine if the given equation 𝑥 squared plus 𝑦 squared plus 𝑧 squared plus two 𝑥 minus two 𝑦 minus eight 𝑧 plus 19 is equal to zero describes a sphere. If so, find its radius and center.

In this question, we’re given an equation. We need to determine whether this equation represents a sphere. And if it does represent a sphere, we need to determine the center of this sphere and the radius of this sphere. The easiest way to do this would be to try and write our equation in the standard form for the equation of a sphere. So let’s start by recalling what this means. We recall a sphere centered at the point 𝑎, 𝑏, 𝑐 with a radius 𝑟 will have the following equation in standard form. 𝑥 minus 𝑎 all squared plus 𝑦 minus 𝑏 all squared plus 𝑧 minus 𝑐 all squared is equal to 𝑟 squared.

What we want to do is rewrite the equation given to us in the question as the equation of a sphere in standard form. And if we were able to do this, we would then be able to read off the center of our sphere and the radius of our sphere. To do this, we’re going to start by rearranging the equation given to us. We’re going to write the 𝑥-terms first, then the 𝑦-terms, then the 𝑧-terms. In the standard form for the equation of a sphere, we see we start with 𝑥 minus 𝑎 all squared. However, in our equation, we have 𝑥 squared plus two 𝑥. To write 𝑥 squared plus two 𝑥 in this form, we’re going to need to complete the square.

We recall to complete the square, we want to write these two terms in the form 𝑥 plus some constant all squared. And to find the constant, we need to halve the coefficient of 𝑥. This is because 𝑥 plus one all squared is equal to 𝑥 squared plus two 𝑥 plus one. If we want this to be equal to 𝑥 squared plus two 𝑥, we need to subtract one from both sides of our equation. Therefore, by completing the square, we’ve shown that 𝑥 squared plus two 𝑥 is equal to 𝑥 plus one all squared minus one. We’re then going to want to do exactly the same thing for our next two terms. This time, we halve the coefficient of 𝑦 to get negative one. And if we distribute the square over our parentheses, we get 𝑦 minus one all squared is equal to 𝑦 squared minus two 𝑦 plus one.

And again, since we want this to be equal to 𝑦 squared minus two 𝑦, we need to subtract one from both sides of our equation, giving us that 𝑦 squared minus two 𝑦 is equal to 𝑦 minus one all squared minus one. Therefore, by completing the square, we were able to rewrite the two terms 𝑦 squared minus two 𝑦 as 𝑦 minus one all squared minus one. Finally, we’re going to need to complete the square on our 𝑧-terms. Once again, inside our parentheses, we’re going to need to halve the coefficient of 𝑧, which is negative eight. This gives us 𝑧 minus four all squared. If we distribute the square over our parentheses, we get 𝑧 squared minus eight 𝑧 plus 16. So to make this equal to 𝑧 squared minus eight 𝑧, we’re going to need to subtract 16 from both sides of our equation. This gives us that 𝑧 squared minus eight 𝑧 is equal to 𝑧 minus four all squared minus 16.

So by completing the square, we were able to rewrite our term 𝑧 squared minus eight 𝑧 as 𝑧 minus four all squared minus 16. And in our equation, we still need to add 19 and set this equal to zero. Therefore, by completing the square three times, we were able to rewrite the equation given to us in the question as 𝑥 plus one all squared minus one plus 𝑦 minus one all squared minus one plus 𝑧 minus four all squared minus 16 plus 19 is equal to zero.

And of course we can simplify this. We have negative one minus one minus 16 plus 19. If we evaluate this, we see it’s equal to one. This means we can rewrite the equation given to us in the question in the following form. But remember, in the standard form for the equation of a sphere, our constant is on the other side of the equation. So we’re going to subtract one from both sides of our equation. This gives us 𝑥 plus one all squared plus 𝑦 minus one all squared plus 𝑧 minus four all squared is equal to negative one.

But now, if we said that this was the equation for a sphere, we would have a problem. In this case, the radius would be the number which squares to give us negative one. However, our radius needs to be positive, so this doesn’t make any sense. So it doesn’t seem that this is the equation for a sphere. In fact, we can prove this. On the right-hand side of our equation, we can see we have a negative number. However, on the left-hand side of the equation, we can see all three of our terms are square. This means all three of our terms are greater than or equal to zero. So the left-hand side of our equation is greater than or equal to zero for all values of 𝑥, 𝑦, and 𝑧. However, the right-hand side of our equation is negative.

Therefore, not only is this not the equation of a sphere, there are no solutions to this equation at all. Therefore, to answer the question, does the equation 𝑥 squared plus 𝑦 squared plus 𝑧 squared plus two 𝑥 minus two 𝑦 minus eight 𝑧 plus 19 equals zero describe a sphere, we were able to show that no, this equation does not describe a sphere.

Let’s now see an example where we use the equation of a sphere to find out some information geometrically about our sphere.

Given that 𝐴 is the point zero, four, four and that the line segment 𝐴𝐵 is a diameter of the sphere 𝑥 plus two all squared plus 𝑦 plus one all squared plus 𝑧 minus one all squared is equal to 38, what is the point 𝐵?

In this question, we’re given some information about a sphere. First, we’re told the standard equation of the sphere. We’re also told that the line segment 𝐴𝐵 is a diameter of our sphere and we’re told the coordinates at the point 𝐴. We need to use all of this information to determine the coordinates of point 𝐵. There’s several different methods we could do this. However, usually in problems like this, the easiest method involves starting by writing down all of the information we’re given. To do this, let’s start by recalling the standard form for the equation of a sphere. We recall a sphere of radius 𝑟 centered at the point 𝑎, 𝑏, 𝑐 will have the following equation in standard form. 𝑥 minus 𝑎 all squared plus 𝑦 minus 𝑏 all squared plus 𝑧 minus 𝑐 all squared is equal to 𝑟 squared.

This means if we’re given the equation of a sphere in standard form, we can find its center point 𝑎, 𝑏, 𝑐 and we can also find its radius 𝑟. And we could see the equation given to us in the question is in standard form, so we can use this to find the center and radius of our sphere. There’s two different methods of finding the center point. We could rewrite the expressions inside of our parentheses as the variable minus some constant. However, we could also just find the value of the variable which makes this term equal to zero. So, for example, our value of 𝑎 would be negative two, our value of 𝑏 would be negative one, and our value of 𝑐 would be one. Either method would work; it’s personal preference which one you want to use. Either way, we’ve shown the center of the sphere given to us in the question is the point negative two, negative one, one.

Similarly, we can find the radius of this sphere by taking the square root of 38. The last thing we’re going to want to do is use the fact that the line segment 𝐴𝐵 is a diameter of our sphere and that the coordinates of the point 𝐴 are zero, four, four. And we might want to sketch this information on a sphere. However, it’s not necessary. We can actually just do this on a circle because if the line segment 𝐴𝐵 is a diameter of the sphere, then it’s also a diameter of the circle of the same radius. In either case, the only information we need is the line segment 𝐴𝐵 is a diameter of our sphere, so it’s a straight line passing through the center of our sphere. And we know the coordinates of the point 𝐴 is zero, four, four and the coordinates of our center 𝑐 is negative two, negative one, one.

And we can combine all of this information to find the coordinates of point 𝐵. First, line segment 𝐴𝐶 and line segment 𝐶𝐵 are both radii of our sphere. They’re both going to have length 𝑟. Next, because we know the coordinates of the point 𝐴 and the coordinates of the point 𝐶, we’re able to find the vector from 𝐴 to 𝐶. And we can also see something interesting. This is going to be exactly the same as the vector from 𝐶 to 𝐵 because they have the same magnitude of 𝑟 and they point in the same direction. So let’s use this to find the coordinates of 𝐵. First, we’re going to need to find the vector 𝐀𝐂. And to do this, we need to take the vector 𝐎𝐂 and subtract the vector 𝐎𝐀. This gives us the following expression, and we can subtract this component-wise. This gives us the vector 𝐀𝐂 is the vector negative two, negative five, negative three. We can then add this onto our sketch. And remember, the vector from 𝐶 to 𝐵 is also equal to the vector from 𝐴 to 𝐶.

Now we can find the coordinates of 𝐵 by adding our vector 𝐎𝐂 to the vector 𝐀𝐂. And all we’re saying here is we can get to the point 𝐵 from the center by moving along the vector 𝐀𝐂. This gives us the vector 𝐎𝐁 will be the vector negative two, negative one, one plus the vector negative two, negative five, negative three. And we add these together component-wise to get the vector 𝐎𝐁 is the vector negative four, negative six, negative two. But remember, the question is not asking us for the vector 𝐎𝐁. It’s asking us for the coordinates of the point 𝐵. And of course, the coordinates of the point 𝐵 will just be the components of our vector 𝐁. And this gives us that 𝐵 is the point negative four, negative six, negative two.

Therefore, we were able to show if 𝐴 is the point zero, four, four and the line segment 𝐴𝐵 is a diameter of the sphere 𝑥 plus two all squared plus 𝑦 plus one all squared plus 𝑧 minus one all squared is equal to 38, then the point 𝐵 must have coordinates negative four, negative six, negative two.

Let’s now go over the key points of this video. First, we know that a sphere is a three-dimensional shape where every point on our sphere is a set distance of 𝑟 from the center. And we call this value of 𝑟 the radius of our sphere. And much like in the case with a circle, we were able to find the equation of a sphere. We were able to show that a sphere centered at the point 𝑎, 𝑏, 𝑐, with a radius of 𝑟 will have the following equation. And we call this equation the standard form of the equation of our sphere. The equation is given by 𝑥 minus 𝑎 all squared plus 𝑦 minus 𝑏 all squared plus 𝑧 minus 𝑐 all squared is equal to 𝑟 squared. And given the standard form of the equation of a sphere, we can find both the center of the sphere and the radius of the sphere.

To find the center of our sphere, the easiest way is to find the values of our variables we would substitute in to make each term equal to zero. And to find the radius of this sphere, all we need to do is take the square root of the constant on the right-hand side of the equation.

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