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The Midpoint and Distance Formulas in 3D - Problem 2
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One application of the distance formula is we can use it to find the equation of a sphere. Here is the definition of a sphere. Given a point C in space, and r greater than 0, the set of points that are at a distance r from c is called a sphere. R is the radius, and c is the center.

So let's do a problem where we find an equation. Let's find an equation of a sphere centered at -1, 3, 2 with radius 2. I've drawn that over here. So here is my sphere, here is the center -1, 3, 2. The radius is 2. I've drawn a point, P, with coordinates x, y, and z representing a general point on the surface of the sphere. Now one thing I know about this point is that it's two units away from C. And do the length of this vector CP, is 2.

So let me first write the components of vector CP. They're going to be x minus -1, so x plus 1, y minus 3, and z minus 2. Now the magnitude, the length of this factor as I said is 2. The length of the vector can be calculated by taking the square root of the sum of the squares of these components. So (x plus 1) squared, plus (y minus 3) squared, plus (z minus 2) squared. That equals and then I square both sides.

Squaring both gets rid of the radical, and I get x plus 1, quantity squared, plus y minus 3, quantity squared, plus z minus 2, quantity squared equals 4. This is what the equation of a sphere looks like. Now notice it looks a lot like the equation of a circle. If I covered this part with my hand, this would look like a circle centered at -1, 3 with radius 2. The only difference between the equation of a sphere, and the equation of a circle is the extra term for the z.

Now in general, a sphere centered at a, b, c, with radius r, will have equation x minus a² plus y minus b², plus z minus c² equals r². That's the general equation in center radius form of a sphere.

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