# Intersection of Two Spheres

• Jan 2nd 2010, 03:12 PM
kid funky fried
Intersection of Two Spheres
For quite sometime I have been intrigued by the curve created by the intersection of two spheres with the following properties/conditions:

Let both spheres have the same radius. Let's say a radius of 2.
Let one sphere be centered at (0,0,0) and the other sphere be centered at
(2,2,2).

I think the equations for the two spheres are:
x^2+y^2+z^2=4
and
(x-2)^2+(y-2)^2+(z-2)^2=4

I have seen a method finding the intersection when both spheres have the same coordinates on the y and z axis and differing coordinates on the x axis. This involves combining the two equations and solving for x.
This would yield a circle parallel to the yz plane.

So, here is what I did:
(x-2)^2+(y-2)^2+(z-2)^2=x^2+y^2+z^2
Multiplying through and rearranging I got
x+y+z=3.
I think this is the equation of the plane where the two spheres intersect.

I am not sure what to do next.
I have read that it will be a circle and that there will be two equations describing the curve.(Now that i think about it i guess the intersection of 2 spheres will always be a circle)
I had built a model about 10 years ago out of balsa wood and the curve really did not look circular. But maybe that was due to the inaccuracies of the model.
How do I find an equation that describes this curve.
I looked in my old calculus book but could find nothing. :(

Any help would be greatly appreciated.
• Jan 2nd 2010, 10:16 PM
matheagle
You have x+y+z=3, but you also have the original constraints.

IF you shift the second center away from (2,2,2) this will be easier to observe.

Look at $(x-2)^2+y^2+z^2=4$ and $x^2+y^2+z^2=4$

Then, when you solve for the intersection you have x=1, but NOT all points on the plane x=1.

Letting x=1 into either/both of your orginal constraints you have $y^2+z^2=3$

so the intersection of these two spheres is x=1 and $y^2+z^2=3$

which is a circle in the plane x=1.

I was playing around with the algebra in your case and it seems to be a circle.

Insert z=3-x-y into the sphere $x^2+y^2+z^2=4$ and see what you get.
• Jan 3rd 2010, 12:38 AM
Abu-Khalil
Quote:

Originally Posted by matheagle
I was playing around with the algebra in your case and I'm not confident that you have a circle.

From where i see it, it's the same than having one sphere centered at $\left(2\sqrt{2},0,0\right)$. Just need a rotation.
• Jan 3rd 2010, 01:23 AM
HallsofIvy
Quote:

Originally Posted by kid funky fried
For quite sometime I have been intrigued by the curve created by the intersection of two spheres with the following properties/conditions:

Let both spheres have the same radius. Let's say a radius of 2.
Let one sphere be centered at (0,0,0) and the other sphere be centered at
(2,2,2).

The intersection two spheres, or of any plane with a sphere, is either empty or a circle. In this case, your two spheres each have radius 2 and the distance between their centers is $\sqrt{4+ 4+ 4}= 2\sqrt{3}< 4$ so they intersect in a circle. That circle will line in the plane that perpedicularly bisects the line between their centers. The midpoint of that line is (1, 1, 1) and a vector in the direction of that line is $\vec{i}+ \vec{j}+ \vec{k}$. The equation of that plane is 1(x-1)+ 1(y- 1)+ 1(z- 1)= 0 or x+ y+ z= 3. Since the distance from (2, 2, 2) to (1, 1, 1) is $\sqrt{3}$ and the radius of each sphere is 2, the radius of the circle of intersection is one leg of a right triangle having one leg of length $\sqrt{3}$ and hypotenuse 2: the radius is $\sqrt{(\sqrt{2^2- (\sqrt{3})^2}= 1$. That is, the intersection is a circle with center at (1, 1, 1), radius 1, in the x+ y+ z= 3 plane.

Quote:

I think the equations for the two spheres are:
x^2+y^2+z^2=4
and
(x-2)^2+(y-2)^2+(z-2)^2=4

I have seen a method finding the intersection when both spheres have the same coordinates on the y and z axis and differing coordinates on the x axis. This involves combining the two equations and solving for x.
This would yield a circle parallel to the yz plane.

So, here is what I did:
(x-2)^2+(y-2)^2+(z-2)^2=x^2+y^2+z^2
Multiplying through and rearranging I got
x+y+z=3.
I think this is the equation of the plane where the two spheres intersect.

I am not sure what to due next.
I have read that it will be a circle and that there will be two equations describing the curve.(Now that i think about it i guess the intersection of 2 spheres will always be a circle)
I had built a model about 10 years ago out of balsa wood and the curve really did not look circular. But maybe that was due to the inaccuracies of the model.
How do I find an equation that describes this curve.
I looked in my old calculus book but could find nothing. :(

Any help would be greatly appreciated.
• Jan 3rd 2010, 08:15 PM
kid funky fried
Thanks
Thanks to all. I appreciate the timely response.

For a given curve in space is there a way to find an equation or set of equations to define the curve?
• Jan 4th 2010, 06:40 PM
HallsofIvy
I've been working on this for some time- mostly going down a blind alley!

As I said before, the intersection of two spheres, each of radius 2, with centers at (0, 0, 0) and (2, 2, 2), is a circle of radius 1, with center at (1, 1,1), in the plane x+ y+ z= 3.

As for finding parametric equations for that circle, matheagles remark, "IF you shift the second center away from (2,2,2) this will be easier to observe", gives a good idea. (Abu-Kahlil said much the same thing [his " $2\sqrt{2}$" should have been " $2\sqrt{3}$"] but it was his "Just need a rotation" that sent me down a blind alley looking for rotation matrices!) If the two spheres were centered at (0, 0, 0) and $(2\sqrt{3}, 0, 0)$, which is the same distance from (0, 0, 0), then their intersection is a circle of radius 1 with center at $(\sqrt{3}, 0, 0)$, in the plane $x= \sqrt{3}$, perpendicular to the x-axis. The parametric equations for that circle are easy- $x= \sqrt{3}$, and we can use the "usual" parametric equations for a circle of radius 1, y= cos(t) and z= sin(t).

Now, we need to find a linear transformation that will change one geometric situation into the other (it doesn't have to be linear but that is simplest). That is, a linear transformation that will take the vectors <1, 0, 0>, <0, 1, 0>, and <0, 0, 1> into axes appropriate to this problem.

The line through the center of one circle, perpendicular to its plane, is the x-axis, y= z= 0, while the line through the center of the other, perpendicular to its plane is x= y= z. A vector in that direction is <1, 1, 1> and a unit vector in that direction is $\left<1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}\right>$.

For a second coordinate axis, we need vector perpendicular to that, so in the plane x+ y+ z= 0. Just about any will do, so take z= 0. That makes the equation x+ y= 0 or y= -x so <-1, 1, 0> will do and a unit vector in that direction is $\left<-1/\sqrt{2}, 1/\sqrt{2}, 0\right>$.

The third axis must be perpendicular to both, so we take the cross product of the first two: $\left<1/\sqrt{6}, 1/\sqrt{6}, -2/\sqrt{6}\right>$.

Now we need a matrix that will map <1, 0, 0> into $\left<1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}\right>$, <0, 1, 0> into $\left<-1/\sqrt{2}, 1/\sqrt{2}, 0\right>$, and <0, 0, 1> into $\left<1/\sqrt{6}, 1/\sqrt{6}, -2/\sqrt{6}\right>$.

It is well known that the matrix that maps three basis vectors into three given vectors is the matrix having those given vectors as columns:
$\begin{bmatrix} \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}}\end{bmatrix}$.

Take the product of that matrix with each of $\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix}$, $\begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}$, and $\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$ to see why that works.

To change the parametric equations $x= \sqrt{3}$, y= cos(t), z= sin(t) to this new "basis" and so give equations for the original circle, we multiply
$\begin{bmatrix} \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}}\end{bmatrix}\begin{bmatrix}\sqr t{3} \\ cos(t) \\ sin(t)\end{bmatrix}$ $= \begin{bmatrix} 1-\frac{1}{\sqrt{2}}cos(t)+ \frac{1}{\sqrt{6}}sin(t) \\ 1+ \frac{1}{\sqrt{2}}cos(t)+ \frac{1}{\sqrt{6}}sin(t) \\ 1- \frac{2}{\sqrt{6}}sin(t)\end{bmatrix}$.

That is, parametric equations for the circle forming the intersection of the two spheres are $x= 1- 1/\sqrt{2} cos(t)+ 1/\sqrt{6}sin(t)$, $y= 1+ 1/\sqrt{2} cos(t)+ 1/\sqrt{6} sin(t)$, and $z= 1- 2/\sqrt{6} sin(t)$.

It is easy to see that those do give the correct figure.
$(x-1)^2= 1/2 cos^2(t)- 1/\sqrt{3} sin(t)cos(t)+ 1/6 sin^2(t)$
$(y-1)^2= 1/2 cos^2(t)+ 1/\sqrt{3} sin(t)cos(t)+ 1/6 cos^2(t)$
$(z-1)^2= 2/3 sin^2(t)$

$1/2 cos^2(t)+ 1/2 cos^2(t)= cos^2(t)$, $-1/\sqrt{3} sin(t)cos(t)+ 1/\sqrt{3} sin(t)cos(t)= 0$, and $1/6 sin^2(t)+ 1/6 sin^2(t)+ 2/3 sin^2(t)= sin^2(t)$ so
$(x-1)^2+ (y-1)^2+ (z-1)^2= cos^2(t)+ sin^2(t)= 1$

Also, since $-\sqrt{2} cos(t)+ 1/\sqrt{2}cos(t)= 0$ and $1/\sqrt{6}sin(t)+ 1/\sqrt{6}sin(t)- 2/\sqrt{6}sin(t)= 0$, we have x+ y+ z= 1+ 1+ 1= 3.

The points given by the parametric equations lie on the intersection of that sphere of radius 1 and that plane. Since the center of the sphere, (1, 1, 1) lies on the plane x+ y+ z= 3, there intersection is the circle of radius 1, with center at (1, 1, 1), in the plane x+ y+ z= 3.
• Jan 5th 2010, 06:23 PM
kid funky fried
Thank You
HallsofIvy,

Thank you so much. I sincerely appreciate the amount of time that you have spent on this problem.
I will print your response, take it to work with a couple of my old text books and study until I understand fully.

Best regards,
Kid