The Midpoint and Distance Formulas in 3D - Problem 4

Here is a problem that doesn't directly use the distance formula. But if you recall, before we use the distance formula to derive the equation of the sphere in center radius form.

Well here I have the equation of the sphere that is not in center radius form. I want to find the center and radius of this sphere. So I have to put it in center radius form. The way we do that is completing the square. It's exactly the same process you use for circles.

So here is how it's done. You want to collect the x terms together, x² plus 2x plus and leave a little space. And the y terms; Y² minus 4y, and leave a little space. Then the z terms; z² and 6z, leave a little space equals 11.

So the idea here is that I need to add a number to these two, to make this into a perfect square trinomial. From experience you probably remember that if I add 1, this becomes x plus 1 quantity squared. So that's a perfect square trinomial, factors like these. So what do I add to y² minus 4y? I have to add 4, and then this becomes y minus 2². Remember the trick, to get this number, you take half of -4 which is -2, and square it, 4. Here I need to take half of 6 which is 3, and square that and I get 9. This is z plus z quantity squared.

Now I've added three numbers to the left side, and I added them to the right side, so I really need to do that. I need to add 1, 4, and 9. This becomes 20 plus 5, 25. Well this actually is an equation of a sphere in center radius form. When you have it in this form, you can identify the center radius really easily. The center is given by whatever makes this 0; -1, 2, -3. The radius is the square root of this. Radius 5. It works a lot like circles.

So just remember, if you have an eqaution like this, x² plus y² plus z² plus, and a bunch of other junk, it might be a sphere. Try completing the square, and putting it into center radius form, and see if you can identify this center and radius.