### Video Transcript

Find the equation of the sphere concentric with π₯ squared plus π¦ squared plus π§ squared plus π₯ minus five π¦ plus four π§ equals three, but with twice the radius.

Two spheres are concentric if they have the same center point. So to find the equation weβre looking for, we need to find the center point and radius associated with the equation that weβre given. Unfortunately, the connection between this equation and the associated radius and center point is not immediately obvious. So letβs recall a different form for the equation of a sphere where the connection is readily apparent.

If π₯ naught, π¦ naught, and π§ naught are the coordinates of the center of the sphere and π is its radius, then the so-called standard form for the equation of the sphere is π₯ minus π₯ naught squared plus π¦ minus π¦ naught squared plus π§ minus π§ naught squared equals π squared. If weβre given an equation for a sphere in this form, we can immediately read off the radius as the square root of the right-hand side and the coordinates of the center of the sphere as π₯ naught, π¦ naught, and π§ naught from the left-hand side. So to answer this question, what we need to do is put the equation weβre given in standard form and then read off the information that we need.

Now, observe that in the standard form for the equation, we only have quadratic terms on the left-hand side, but in the equation that weβre given, we have both quadratic and linear terms. So letβs expand the left-hand side of the equation in standard form to see where both the quadratic and the linear terms come from. π₯ minus π₯ naught squared is π₯ squared minus two π₯ naught π₯ plus π₯ naught squared. π¦ minus π¦ naught squared and π§ minus π§ naught squared are the same but with π¦ and π§ substituted for π₯. Our fully expanded standard form now consists of three quadratic terms, three linear terms, and three constant terms on the left-hand side of the equation.

The three quadratic terms in our standard form exactly match the three quadratic terms in the equation weβre given. This means that the three linear terms must also correspond exactly to the three linear terms in that equation. However, unlike the equation in standard form, there are no constant terms on the left-hand side of this equation. All that means, though, is that all of the constants in the equation are collected on the right-hand side, which can be accomplished by subtracting π₯ naught squared, π¦ naught squared, and π§ naught squared from both sides of the equation in standard form.

Remember, though, our goal is to bring this equation into standard form, so we need a term-by-term relationship between this equation and the equation we got from the standard form. As weβve mentioned, the quadratic terms are already taking care of appearing exactly the same way in both equations. For all of the other terms, weβll need values for π₯ naught, π¦ naught, and π§ naught. And we can find these by comparing the linear terms in our equation from standard form to the linear terms in the equation weβre given. This works because linear terms can only combine with other linear terms with the same variable. So we know that negative two π₯ naught π₯ is equal to π₯, negative two π¦ naught π¦ is equal to negative five π¦, and negative two π§ naught π§ is equal to four π§.

To solve these equations, we simply divide both sides of the first equation by negative two π₯, the second equation by negative two π¦, and the third equation by negative to π§. This gives us that π₯ naught is negative one-half, π¦ naught is five-halves, and π§ naught is negative two. To find the radius, recall that three is the collection of all of the constant terms in the equation, which we accomplish by subtracting π₯ naught squared, π¦ naught squared, and π§ naught squared from both sides. So three is π squared minus π₯ naught squared minus π¦ naught squared minus π§ naught squared.

Now, π₯ naught squared is negative one-half squared or one-quarter, π¦ naught squared is five-halves squared or twenty-five quarters, and π§ naught squared is negative two squared, which is four. Negative one-quarter minus twenty-five quarters minus four is negative 10 and one-half. So since three is π squared minus 10 and one-half, adding 10 and one-half to both sides gives us π squared is 13 and one-half. So now we know both the center and the square of the radius of the original sphere that we were given. And all we need to do is translate this into the equation of a sphere with the same center but twice the radius. If we call the new radius capital π
, then we have that capital π
is equal to two times lowercase π, the old radius.

Squaring both sides gives us capital π
squared equals two times lowercase π squared, which is equal to four times lowercase π squared. But we have a value for lowercase π squared. Itβs just 13 and one-half. Four times 13 and one-half is 54. So the square of the new radius is 54. Now we have the center of our new sphere, which is the same as the center of our old sphere, and the square of the radius of our new sphere, which gives us everything we need to find the equation of the new sphere in standard form. Using these values, the equation of our new sphere is π₯ plus one-half squared plus π¦ minus five-halves squared plus π§ plus two squared equals 54.