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Numerade Educator

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Problem 47 Easy Difficulty

Find the distance between the spheres $ x^2 + y^2 + z^2 = 4 $ and $ x^2 + y^2 + z^2 = 4x + 4y + 4z - 11 $.

Answer

$2 \sqrt{3}-3$

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Video Transcript

Let's find the distance between these two spheres. The Sphere X squared plus y squared plus C squared equals four and X squared plus y squared plus C squared equals four x plus four y plus four Z minus 11. What's that distance going to look like? So let me do this as projection so I don't have a cluttered diagram. So is that I have two fears. So here's once here and here's another. We're looking for this distance between that. Well, that distance can be found by looking at the distance between the center of the two spheres minus the radius of each sphere, and that will give us this distance. In between that we're looking for, that's what we're going to do. We're going to find the distances between the centres of each sphere and then subtract the radius of X fear and that will give us the distance between them. Well, the first year has its center at the origin. We don't need to do any work to see that, and we can see that it's radius is to to find the center and radius of the second sphere. We're going to have to do some completing the square. So let's rewrite our expression, bringing all of the variable terms to the left hand side I X squared minus four x plus the time to complete the square plus y squared minus four. Why, plus a term to complete the square plus Z squared minus four z plus a term to complete the square equals negative 11 plus the same three terms. But I had to after the left hand side that's actually add to the right hand side to balance the equation. And so, for all of these, half of negative, too half of negative four is negative. Two and negative two squared is for so to complete the square for all of these, I have to add. Four. You have to do that on the right hand side as well. That's going to give me X minus two squared plus why minus two square plus Z minus two square and negative 11 plus 12 is one. So this has a center at 22 to, and the radius is one because once where it is one. So let's find the distances between the centres and then we will subtract three from that because the two Radio I after three so two and one. So find this distance and subtract three, and that will be the distance between the two spheres. So distance between centres is a square root of subtracting X coordinates to minus zero square, subtracting like award. It's to minus zero square, subtracting Z coordinates to minus zero. Swear so I get to the square root of two square plus two squared Plus two square, which is the square root of four plus four plus four to get the square root of 12. 12 is four times three. Spirit of four is two seconds. Bring out the to and say this is two times the square root of three. And so the distance between the spheres it's gonna be the distance between the centres. Two times World of free minus. That's some of the radio two plus one, so I get to time to screw to three minus three, and there's no real way to simplify this. If you want a decimal expression, you can get out your calculator, but I don't think that's needed to time. Just for three minus three is a sufficient expression to represent the distance between those two Sears