The intersection of the sphere $\displaystyle x^2 + y^2 + z^2 = 6$ with the plane $\displaystyle x - 2y + 3z = 7$ is a circle. What is the z-coordinate of the center of the circle?

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- Oct 11th 2010, 09:36 AMSyNtHeSiSIntersection of plane with sphere
The intersection of the sphere $\displaystyle x^2 + y^2 + z^2 = 6$ with the plane $\displaystyle x - 2y + 3z = 7$ is a circle. What is the z-coordinate of the center of the circle?

- Oct 11th 2010, 10:27 AMJhevon
From the plane, $\displaystyle \displaystyle x - 2y + 3z - 1 = 6$, hence

$\displaystyle \displaystyle x^2 + y^2 + z^2 = x - 2y + 3z - 1$

Now, get all the variables on one side, do all that completing the square thing, and then you can find the z-coordinate for the (center of the) circle. - Oct 11th 2010, 10:40 AMSyNtHeSiS
Thanks I understand what you did but I am wondering if you equate the plane and sphere, wouldnt this find the points of intersection? I dont understand why the z coordinate would give you the center of the circle.

- Oct 11th 2010, 10:47 AMJhevon
yes, intersection means being at the same point(s), so you set both equations equal to find the common points.

well, there are many points on the circle, and potentially an infinite amount of z-points, meaning, all the coordinates on the circle have a different z-coordinate. the problem makes sense to me if you state a unique point, which makes sense to be the center.

if that's not what they want (did you type the question__exactly__as it appeared?), then simply solve for z in the last equation i gave you - Oct 11th 2010, 02:26 PMSyNtHeSiS
Sorry but I forgot to add that you must find the z-coordinate of the center of the circle. How would you go about it now?

- Oct 11th 2010, 03:02 PMJhevon
Haha, that's what i've been saying. what i told you to do you can get that. by completing the square, you will be able to write the equation as

$\displaystyle \displaystyle (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2$

$\displaystyle \displaystyle c$ is the number you want - Oct 12th 2010, 03:52 AMearboth
1. Obviously you have some doubts whether Jhevon's solution is valid or not. It is but ...

2. The**equation**you got from Jhevon describes**a sphere**. This sphere is the smallest which fits exactly into the circle which contains the points of intersection between the plane and the original sphere.

3. Therefore the plane cuts the 2nd sphere at it's equator and therefore the center of the sphere is simultaneously the center of the circle you are looking for.

4. In the attached sketch I haven't drawn the lower halfsphere so you have a better "insight". - Oct 12th 2010, 03:57 AMAlso sprach Zarathustra
Hey

**earboth**!

Can you please tell in which program you used to get that nice output? - Oct 12th 2010, 08:46 AMearboth
1. I use Derive 6.0. This program isn't available any more.

2. If I were you I would use Maxima. Have a look here: Maxima, a Computer Algebra System