# Calc 2 equations of sphere

1. Apr 23, 2010

### somebodyelse5

Find the equation of a sphere if one of its diameters has endpoints: (-1, -5, -8) and (3, -1, -4)

Ok, heres what I have so far, I cant find the radius and im not sure if the rest of the equation is correct. Havin some issues with this one.

(x-1)^2+(y+2)^2+(z+2)^2-?=0

2. Apr 23, 2010

### gabbagabbahey

Hint: You are given the endpoints of one of the diameters, shouldn't the center of the sphere be halfway between the two endpoints?

3. Apr 23, 2010

### somebodyelse5

So I subtract my vectors, and get (4,-4,-4) correct?

Then I divide that vector by 2 and get (2, -2, -2) Correct?
So then my radius should be 2

So that means my equation should be
(x-2)^2+(y+2)^2+(z+2)^2-2^2=0

But when i enter that its wrong.

4. Apr 23, 2010

### gabbagabbahey

First, double check your negative signs. Second, what quantity does subtracting the two position vector really give you?

5. Apr 23, 2010

### somebodyelse5

Ok, Im having a really difficult time doing the simplest part of this problem.
So, by subtracting the two points, I get the diameter of the sphere, then dividing it by 2 gives me the radius.

Redid it and got a difference of (4,4,4) and a radius of (2,2,2)

I think i have an idea of where Im going wrong, by dividing the diameter by 2 i do get the radius, but that is not necessarily the center point. How would I go about finding the center point?

6. Apr 23, 2010

### gabbagabbahey

No, diameter is a distance (scalar)...subtracting two position vectors gives you a vector...specifically the vector from the first endpoint to the second endpoint....What does dividing that vector by 2 give you? (draw a picture if you aren't sure!)

That looks better

7. Apr 23, 2010

### somebodyelse5

Got this one figured out also. Answer is (x-1)^2+(y+3)^2+(z+6)^2-3.464^2

And used the distance formula, which i forgot about, to solve for the distance and then divided that by 2 to get the radius. Then I basically took the average of each x y and z point to find the center.

Thanks for you help!! Glad i found this site
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