Video Transcript
Does the equation two 𝑥 squared plus two 𝑦 squared plus two 𝑧 squared plus four 𝑥 plus four 𝑦 plus four 𝑧 minus 44 equals zero describe a sphere? If so, find its radius and center.
In this question, we’re given an equation, and we’re asked to determine if this equation describes a sphere. And we know that a sphere is a shape in three dimensions which represents all the points a set distance away from its center. We call this the radius of the sphere. If this equation does indeed represent a sphere, we need to determine the radius and the center from this equation.
To help us answer this question, let’s start by recalling how we can represent a sphere with an equation. We know that a sphere centered at the point 𝑎, 𝑏, 𝑐 with a radius of 𝑟, which must be positive, has the equation 𝑥 minus 𝑎 all squared plus 𝑦 minus 𝑏 all squared plus 𝑧 minus 𝑐 all squared is equal to 𝑟 squared. This is known as the standard form of the equation of a sphere. All spheres can be represented in standard form. And we can find both the center of the sphere and its radius from its equation given in standard form.
Since all spheres can be represented in standard form, to answer this question, we need to determine if we can rewrite the equation given to us in the question in this standard form. To try and rewrite this equation in standard form, we need to notice in the standard form of the equation of a sphere, we have three binomials which include just the variable 𝑥, 𝑦, and 𝑧, respectively. This means we’ll want to rewrite this equation in terms of three binomials.
However, this immediately gives us a problem. If we were to try and expand the term 𝑥 minus 𝑎 all squared by using the FOIL method or binomial expansion, we would get 𝑥 squared minus two 𝑎𝑥 plus 𝑎 squared. We can see the coefficient of 𝑥 squared is equal to one. However, in the equation, we’re given the coefficient of 𝑥 squared is two. What this tells us is in the standard form of the equation of a sphere, the coefficients of 𝑥 squared, 𝑦 squared, and 𝑧 squared will be equal to one. So the first thing we’ll do is rewrite the equation we’re given so that the coefficients of these three terms is equal to one. We’ll do this by dividing through by two.
Multiplying both sides of this equation through by one-half means we need to multiply each term by one-half. This is the same as dividing each term through by two. We get the equation 𝑥 squared plus 𝑦 squared plus 𝑧 squared plus two 𝑥 plus two 𝑦 plus two 𝑧 minus 22 is equal to zero. We now want to try and write this equation in the standard form for the equation of a sphere.
In the standard form for the equation of a sphere, all of the 𝑥-terms are contained in the square of the binomial 𝑥 minus 𝑎 all squared. Therefore, in our equation, we’re going to want to group 𝑥 squared and two 𝑥 into the same form. One way of doing this is to complete the square on the terms 𝑥 squared and two 𝑥. Remember, to complete the square when the coefficient of 𝑥 squared is equal to one, we start by halving the coefficient of 𝑥, which in this case is two, to give us a value of one.
If we were to then square 𝑥 plus one, we could distribute the exponent by using the FOIL method or binomial expansion, and we would get 𝑥 squared plus two 𝑥 plus one. And this is almost exactly the same as what we started with, 𝑥 squared plus two. However, we’ve added an extra constant of one. Therefore, if we subtract one from 𝑥 plus one all squared, it will be equal to 𝑥 squared plus two 𝑥. This means that we’ve shown by using completing the square 𝑥 squared plus two 𝑥 is equal to 𝑥 plus one all squared minus one.
We can use this to rewrite the terms 𝑥 squared plus two 𝑥 in our equation. We can replace 𝑥 squared plus two 𝑥 with 𝑥 plus one all squared minus one. To write this equation in the standard form of the equation of a sphere, we’re going to need to do the same with our 𝑦- and 𝑧-terms. Let’s do the same with 𝑦 squared plus two 𝑦.
Normally, we would need to complete the square with these two terms again. However, we can see this is exactly the same expression we have with 𝑥. However, we’re now using the variable 𝑦. In other words, we can just replace 𝑥 with 𝑦 in this expression to complete the square with 𝑦. 𝑦 squared plus two 𝑦 will be equal to 𝑦 plus one all squared minus one. We can then use this to rewrite the terms 𝑦 squared plus two 𝑦 in our equation. We replace 𝑦 squared plus two 𝑦 with 𝑦 plus one all squared minus one.
Finally, we’re going to want to do the same with our 𝑧-terms. And we can see that this is 𝑧 squared plus two 𝑧. Once again, to write this in the standard form of the equation of a sphere, we could use completing the square. However, we can also notice this is exactly the same as the expressions we’ve already found. We just need to replace 𝑥 or 𝑦 with 𝑧. We know that 𝑧 squared plus two 𝑧 will be equal to 𝑧 plus one all squared minus one. Therefore, we can replace 𝑧 squared plus two 𝑧 in our equation with 𝑧 plus one all squared minus one.
Finally, the equation we’re given tells us if we subtract 22 from this, it must be equal to zero. This is now almost in the standard form for the equation of a sphere. We just need to write the constant on the right-hand side of the equation. To do this, we notice that negative one minus one minus one minus 22 is equal to negative 25. So we’ll add 25 to both sides of our equation. Doing this, we get the equation 𝑥 plus one all squared plus 𝑦 plus one all squared plus 𝑧 plus one all squared is equal to 25.
Finally, in the standard form for the equation of a sphere, we usually write the constant in the form 𝑟 squared, where 𝑟 is positive because it represents the radius of our sphere. We can do this in this case by remembering that 25 is equal to five squared. Therefore, we’ve shown the equation given to us in the question is equivalent to the equation 𝑥 plus one all squared plus 𝑦 plus one all squared plus 𝑧 plus one all squared is equal to five squared. This is now exactly in the standard form for the equation of a sphere of radius five.
But remember, the question also wants us to find the center of our sphere. We do this by solving each of the individual binomials equal to zero. Solving the first binomial equal to zero, we get that 𝑥 plus one should be equal to zero. We know this is true when 𝑥 is equal to negative one. This means the 𝑥-coordinate of the center of our sphere is negative one. We can do the same with our other two binomials. We get that 𝑦 is equal to negative one and 𝑧 is equal to negative one.
Therefore, we’ve shown that the center of the sphere is the point negative one, negative one, negative one. And hence we’ve answered the question “does the equation two 𝑥 squared plus two 𝑦 squared plus two 𝑧 squared plus four 𝑥 plus four 𝑦 plus 𝑧 minus 44 equals zero describe a sphere?” We’ve shown that yes, it does describe a sphere. Its radius is five, and its center is at negative one, negative one, negative one.
