Question Video: Finding the Equation of a Sphere Given the Coordinates of Two Points on Its Diameter

Find the equation of the sphere with 𝐴 = (9, βˆ’6, 1) and 𝐡 = (βˆ’16, βˆ’12, 2) as endpoints of a diameter.

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Video Transcript

Find the equation of the sphere with 𝐴 equal to the point nine, negative six, one and 𝐡 equal to the point negative 16, negative 12, two as endpoints of a diameter.

In this question, we are asked to find the equation of a sphere. And to help us do this, we’re given two points which, we’re told, lie as endpoints of one of the diameters of the sphere. To answer this question, let’s start by recalling the equation of a sphere. We know that a sphere centered at the point π‘Ž, 𝑏, 𝑐 with a radius π‘Ÿ which is positive will have the equation π‘₯ minus π‘Ž all squared plus 𝑦 minus 𝑏 all squared plus 𝑧 minus 𝑐 all squared is equal to π‘Ÿ squared. This is called the standard form of the equation of a sphere.

And to find this equation for our sphere, all we need to know is its center point and its radius. We need to find both of these pieces of information from the two points which we’re told lie as endpoints of the diameter of our sphere. We can find these by recalling the diameter of a sphere passing through the points 𝐴 and 𝐡 on the surface of our sphere is a line segment passing through the center of our sphere. Let’s call this center 𝐢. Then the line segment 𝐢𝐴 and the line segment 𝐢𝐡 will both be radii of our sphere. So they will be equal in length.

This means the center of our sphere will always be the midpoint of the two endpoints of our diameter. Therefore, we can find the center of our sphere by finding the midpoint of the line segment 𝐴𝐡. And we can find the radius of the sphere by finding the length of the line segment 𝐢𝐴 or the length of the line segment 𝐢𝐡.

Let’s start by finding the coordinates of the center of our sphere. To do this, we need to recall to find the midpoint of two points in space, we need to take the average of their coordinates. In other words, the midpoint between the two points π‘₯ sub one, 𝑦 sub one, 𝑧 sub one and the point π‘₯ sub two, 𝑦 sub two, 𝑧 sub two is given by the point π‘₯ sub one plus π‘₯ sub two all over two, 𝑦 sub one plus 𝑦 sub two all over two, 𝑧 sub one plus 𝑧 sub two all over two. This just means to find the coordinates of the center of our sphere 𝐢, we need to take the average of the coordinates of 𝐴 and 𝐡.

𝐢 will be the point nine plus negative 16 all over two, negative six plus negative 12 all over two, one plus two all over two. And if we evaluate each of the expressions we have for each coordinate, we get that 𝐢 is the point negative seven over two, negative nine, three over two. This means we found the coordinates of the center of our sphere. All we need to find now is the radius of this sphere to find its equation in standard form.

There are many different ways of finding this radius. We’re going to find one-half the distance between the two points 𝐴 and 𝐡. We’ll start by just finding length of our diameter. That’s the distance between the two points 𝐴 and 𝐡. To do this, we recall the distance between the two points π‘₯ sub one, 𝑦 sub one, 𝑧 sub one and π‘₯ sub two, 𝑦 sub two, 𝑧 sub two is given by the square root of π‘₯ sub one minus π‘₯ sub two all squared plus 𝑦 sub one minus 𝑦 sub two all squared plus 𝑧 sub one minus 𝑧 sub two all squared.

Substituting the coordinates of the points 𝐴 and 𝐡 into this formula will give us the length of the diameter of our sphere. We’ll call this 𝑑. It’s equal to the square root of nine minus negative 16 all squared plus negative six minus negative 12 all squared plus one minus two all squared. And of course, remember 𝑑 will be equal to two multiplied by π‘Ÿ. It’s twice the radius.

Evaluating the expression inside of our square root symbol gives us that two π‘Ÿ will be equal to the square root of 662. We can then find the exact value of the radius by dividing both sides of this equation by two. We get π‘Ÿ is equal to one-half root 662. Therefore, we found the radius of our sphere. And we already know the center of our sphere. This means we can find an equation for our sphere in standard form. We can also see in this equation for our sphere in standard form we need an expression for π‘Ÿ squared. We can find an expression for π‘Ÿ squared by squaring our equation for π‘Ÿ.

Squaring both sides of this equation gives us that π‘Ÿ squared will be equal to one-half root 662 all squared. We can evaluate this by distributing the square over our product. This will be equal to one-half squared multiplied by root 662 squared, which is, of course, equal to one-quarter multiplied by 662.

Finally, we can simplify this expression for π‘Ÿ squared to 331 over two. All we need to do now is substitute the coordinates of the center of our sphere and our expression for π‘Ÿ squared into the standard form of the equation of a sphere. Doing this gives us our final answer. The equation of a sphere with the point 𝐴 nine, negative six, one and 𝐡 negative 16, negative 12, two as endpoints of a diameter of the sphere is given by π‘₯ plus seven over two all squared plus 𝑦 plus nine all squared plus 𝑧 minus three over two all squared is equal to 331 over two.

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