# Thread: Intersection of plane with sphere

1. ## Intersection 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?

2. Originally Posted by SyNtHeSiS
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 circle?
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.

3. 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.

4. Originally Posted by SyNtHeSiS
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.
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

5. 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?

6. Originally Posted by SyNtHeSiS
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?
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

7. Originally Posted by SyNtHeSiS
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.
Originally Posted by Jhevon
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
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".

8. Hey earboth!

Can you please tell in which program you used to get that nice output?

9. Originally Posted by Also sprach Zarathustra
Hey earboth!

Can you please tell in which program you used to get that nice output?
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

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### intersection plane and sphere

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