Question Video: Finding the Equation of a Sphere That Passes through Three Points given Their Coordinates

Find the equation of a sphere that passes through the points 𝐴(9, 0, 0), 𝐡(3, 13, 5), and 𝐢(11, 0, 10), given that its center lies on the 𝑦𝑧-plane.

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Video Transcript

Find the equation of a sphere that passes through the points 𝐴 nine, zero, zero; 𝐡 three, 13, five; and 𝐢 11, zero, 10, given that its center lies on the 𝑦𝑧-plane.

One of the most natural ways to express the equation of a sphere is in the so-called standard form. In general, we write the standard form of the equation of a sphere as π‘₯ minus π‘₯ naught squared plus 𝑦 minus 𝑦 naught squared plus 𝑧 minus 𝑧 naught squared equals π‘Ÿ squared. π‘₯ naught, 𝑦 naught, and 𝑧 naught are the coordinates of the center of the sphere and π‘Ÿ is its radius. The sphere itself is then the collection of all the points with coordinates π‘₯, 𝑦, and 𝑧 that satisfy this equation.

What makes this form so natural and useful is that specifying the center of a sphere and its radius is enough to completely determine the entire sphere. And given the equation of a sphere in standard form, we merely need to look at the equation to determine the coordinates of the center of the sphere and its radius.

In our case, though, we don’t know the radius of the sphere and all we know about the center is that it lies on the 𝑦𝑧-plane. We are, however, given three points that lie on the sphere. So to find the equation of the sphere, we’ll take the coordinates of each of these points, plug them into our standard form for the equation of a sphere, and use this to arrive at a system of equations that we can then solve to determine the center of the sphere and its radius, which will then give us the general equation of the sphere. We actually already know one of the coordinates of the center of the sphere. Because the center of the sphere lies in the 𝑦𝑧-plane, the π‘₯-coordinate is zero.

Let’s now plug the points 𝐴, 𝐡, and 𝐢 into our standard form equation. For 𝐴, π‘₯ is nine, 𝑦 is zero, and 𝑧 is also zero. So we have nine minus zero squared because π‘₯ naught is zero plus zero minus 𝑦 naught squared plus zero minus 𝑧 naught squared equals π‘Ÿ squared. Nine minus zero is just nine, zero minus 𝑦 naught is negative 𝑦 naught, and zero minus 𝑧 naught is negative 𝑧 naught. Nine squared is 81. Negative 𝑦 naught squared is negative 𝑦 naught times negative 𝑦 naught, which is positive 𝑦 naught squared. And likewise, negative 𝑧 naught squared is positive 𝑧 naught squared. So 81 plus 𝑦 naught squared plus 𝑧 naught squared equals π‘Ÿ squared.

Now we’ll do the same calculation for point 𝐡. We have π‘₯ is three, 𝑦 is 13, and 𝑧 is five. The only easy simplification we can make is that three minus zero squared is nine. For now, we’ll leave 13 minus 𝑦 naught squared and five minus 𝑧 naught squared as they are. So our second equation is nine plus 13 minus 𝑦 naught squared plus five minus 𝑧 naught squared equals π‘Ÿ squared. Our last equation will come from point 𝐢 which has π‘₯ equals 11, 𝑦 equals zero, and 𝑧 equals 10. Now we have 11 minus zero squared, which is 11 squared or 121; zero minus 𝑦 naught squared, which we know is just 𝑦 naught squared; and 10 minus 𝑧 naught squared, which we won’t expand for the time being. So our third equation is 121 plus 𝑦 naught squared plus 10 minus 𝑧 naught squared equals π‘Ÿ squared.

Now we have a system of three equations in three variables: 𝑦 naught, 𝑧 naught, and π‘Ÿ. To solve this system, we’ll work one variable at a time. Note that all three expressions on the left-hand side are equal to the same expression on the right-hand side, namely, π‘Ÿ squared. So we can create an equation without π‘Ÿ in it by equating any two expressions from the left-hand side of these equations. Also note that in the first and third equations, the only appearance of 𝑦 naught is in the single term 𝑦 naught squared. So by equating the left-hand sides of the first and third equations, we should be able to eliminate not only π‘Ÿ, but also 𝑦 naught, which will leave us with a single equation in the single unknown 𝑧 naught.

So we have 81 plus 𝑦 naught squared plus 𝑧 naught squared equals 121 plus 𝑦 naught squared plus 10 minus 𝑧 naught squared. And we know this equality holds because both expressions are equal to the same thing, π‘Ÿ squared. Now we need to simplify. We’ll start by subtracting 81 and 𝑦 naught squared from both sides. 81 minus 81 is zero, as is 𝑦 naught squared minus 𝑦 naught squared. So on the left-hand side, we’re just left with 𝑧 naught squared. On the right-hand side, 121 minus 81 is 40 and 𝑦 naught squared minus 𝑦 naught squared is again zero. So this leaves us with 𝑧 naught squared equals 40 plus 10 minus 𝑧 naught squared.

Now look what we have: exactly what we wanted, a single equation with the single unknown 𝑧 naught. To solve for 𝑧 naught, we need to expand 10 minus 𝑧 naught squared on the right-hand side. This gives us 𝑧 naught squared equals 40 plus 100 minus 20𝑧 naught plus 𝑧 naught squared. This has actually worked out more nicely than we may’ve originally thought. The fact that we have a 𝑧 naught squared on one side suggests that we have a quadratic equation. But we have the same 𝑧 naught squared on the other side of the equation. So, in fact, this equation is linear and easy to solve.

To solve this equation, we’ll subtract 𝑧 naught squared from both sides and add 20𝑧 naught to both sides. On both sides of the equation, 𝑧 naught squared minus 𝑧 naught squared is zero. This leaves us with 20𝑧 naught on the left-hand side. And on the right-hand side, 40 plus 100 is 140 and negative 20𝑧 naught plus 20𝑧 naught is zero. So 20𝑧 naught equals 140 and dividing both sides by 20, 𝑧 naught equals seven. Now that we have a value for 𝑧 naught, the only two variables left are π‘Ÿ and 𝑦 naught.

To find 𝑦 naught, we’ll equate the expression on the left-hand side of the first and second equation because we know that equating the first and third equation eliminates 𝑦 naught from the calculation. But we need 𝑦 naught in the calculation to find its value. So we have 81 plus 𝑦 naught squared plus 𝑧 naught squared equals nine plus 13 minus 𝑦 naught squared plus five minus 𝑧 naught squared, where again we know this equality holds because both sides are equal to π‘Ÿ squared.

To turn this equation into a single equation with a single variable, all we need to do is plug in 𝑧 naught equals seven. Now we have a single equation with a single variable and all that’s left to do is solve for 𝑦 naught. Let’s start by simplifying and expanding all of the squares that we can. For starters, seven squared is 49. Five minus seven is negative two and negative two squared is positive four. Finally, we expand 13 minus 𝑦 naught squared to 169 minus 26𝑦 naught plus 𝑦 naught squared. Collecting all the numerical terms, on the left-hand side, we have 81 plus 49, which is 130. So the left-hand side is 130 plus 𝑦 naught squared. And on the right-hand side, we have nine plus 169 plus four, which is 182. So the right-hand side is 182 minus 26𝑦 naught plus 𝑦 naught squared.

We see again that what originally looked like a quadratic equation is in fact linear because we have the same squared term on both sides. Subtracting 𝑦 naught squared from both sides gives us 130 equals 182 minus 26𝑦 naught. Now we add 26𝑦 naught and subtract 130 from both sides, which gives us 26𝑦 naught equals 52. It follows then by dividing both sides by 26 that 𝑦 naught is two. The last quantity we need to determine is π‘Ÿ or, more precisely, π‘Ÿ squared. To find this, we’ll simply substitute seven for 𝑧 naught and two for 𝑦 naught in our first equation. So 81 plus two squared plus seven squared is π‘Ÿ squared. That is, 81 plus four plus 49 is π‘Ÿ squared and π‘Ÿ squared is 134.

We now have all four quantities that we need to express the equation of our sphere in standard form. The equation of our sphere in standard form is π‘₯ minus zero squared plus 𝑦 minus two squared plus 𝑧 minus seven squared equals 134. Let’s express this equation in another form where we will expand out all the squares and have the right-hand side of the equation equals zero. π‘₯ minus zero squared is π‘₯ squared. 𝑦 minus two squared is 𝑦 squared minus four 𝑦 plus four. 𝑧 minus seven squared is 𝑧 squared minus 14𝑧 plus 49. And we’ve set the right-hand side equal to zero by subtracting 134 from both sides.

Just to tidy up this expression a little bit, we’ll collect the three quadratic terms together, then the two linear terms. And finally, collecting the constant terms, four plus 49 minus 134 is negative 81, and this whole expression equals zero. π‘₯ squared plus 𝑦 squared plus 𝑧 squared minus four 𝑦 minus 14𝑧 minus 81 equals zero is the equation of a sphere that we’re looking for. And we got this equation from the equation of the sphere in standard form, which we found by solving for the coordinates of the center of the sphere and the square of its radius.

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