4
$\begingroup$

The three equations of spheres are given

$(x-x_{1})^2+(y-y_{1})^2+(z-z_{1})^2=a^2$

$(x-x_{2})^2+(y-y_{2})^2+(z-z_{2})^2=b^2$

$(x-x_{3})^2+(y-y_{3})^2+(z-z_{3})^2=c^2$

How do I find $(x,y,z)$ analytically?

$\endgroup$
5
  • $\begingroup$ $(x-x_{3})^2+(y-y_{3})^2+(z-z_{3})^2=a^2$ or $c^2$? $\endgroup$
    – ulead86
    Nov 11, 2013 at 9:46
  • $\begingroup$ @ulead86 yes. Thanks $\endgroup$
    – Harry
    Nov 11, 2013 at 9:47
  • 1
    $\begingroup$ $S_1 \bigcap S_2 \bigcap S_3 = (S_1 \bigcap \ S_2) \bigcap (S_2 \bigcap S_3) \bigcap (S_3 \bigcap S_1)$. This is essentially intersection of three circles of (possibly) different radii in $\mathbb R^3$. This can be a circle, two distinct points, one point and null set. $\endgroup$
    – ARi
    Nov 11, 2013 at 9:55
  • $\begingroup$ @ulead86 in your solution when you simplified, $z$ term is missing $\endgroup$
    – Harry
    Nov 11, 2013 at 10:15
  • $\begingroup$ To state ARi's point in a different way, the intersection of 2 spheres with distinct centers is a circle with complex radius. The intersection of a circle and sphere with 2 distinct centers is 2 (not necessarily distinct) complex points. $\endgroup$
    – DanielV
    Nov 12, 2013 at 10:04

2 Answers 2

3
$\begingroup$

This is probably not the simplest possible analytic answer.

First, I'm going to choose some easier variables to use. Let $V$ ($V$ is your x,y,z) be the axis variables and $C$ be the centers of 3 spheres:

$$(V_{1} - C_{1,1})^2 + (V_{2} - C_{1,2})^2 + (V_{3} - C_{1,3})^2 = r_1^2 \tag {Sphere1 of V}$$ $$(V_{1} - C_{2,1})^2 + (V_{2} - C_{2,2})^2 + (V_{3} - C_{2,3})^2 = r_2^2 \tag {Sphere2 of V}$$ $$(V_{1} - C_{3,1})^2 + (V_{2} - C_{3,2})^2 + (V_{3} - C_{3,3})^2 = r_3^2 \tag {Sphere3 of V}$$


To begin solving this, first find the intersection of 2 spheres.

First realize that the intersection of 2 spheres is a circle. It could be a circle with real, zero, or complex radius, but it is still a circle. So first we will solve the intersection of 2 spheres analytically.

Choose an affine change of basis $V \rightarrow W$ with new axises $E$ and origin $\Omega$ that will result in the spheres being placed such that:

1) Sphere 1 and 2's centers are on the new $W_1$ axis.
2) The circle that results from their intersection is on the $W_1 = 0$ plane (by choice of location of the new origin).
By only choosing the first basis vector and the origin, these 2 conditions are enough to solve the sphere-sphere problem, but for convenience later we also want:
3) Sphere 3's center is located on the new y axis.
The final condition defines where other 2 basis vectors will be.

The x-axis will pass through the center of the spheres: $$E_1 = norm(C_2 - C_1)$$ The origin is also on this axis: $$\Omega = (C_2 - C_1)\cdot n_1 + C_1$$ The new Y axis passes through sphere 3: $$E_2 = norm(C_3 - \Omega)$$ $$E_3 = E_1 \times E_2$$

This defines an affine change of basis : $$V = W \begin{bmatrix} E_1 \\ E_2 \\ E_3 \end{bmatrix} + \Omega \tag{Transform 1}$$

Let $n$ be the distance between $C_1$ and $C_2$, where $n_1$ is the distance from the relocated Sphere 1 from the new origin and $n_2$ is the distance of the new relocated Sphere 2 from the origin. Last thing before translating the spheres, solve for $n_1$ and $n_2$, (and remember $r_x$, the radius of the circle-of-intersection, because we will definitely need it later ) :

$$n = |C_2 - C_1|$$ $$n = n_1 + n_2$$ $$r_1^2 - n_1^2 = r_2^2 - n_2^2 = r_X^2$$

Solves to

$$n_1 = \frac {n^2 - r_2^2 + r_1^2} {2 n} $$ $$n_2 = \frac {n^2 + r_2^2 - r_1^2} {2 n} $$ $$r_x^2 = - \frac {\left(n^2 - (r_1 - r_2)^2\right)\left(n^2 - (r_1 + r_2)^2\right)} {4 n^2} $$

Now when we put Sphere 1, Sphere 2, and Sphere 3 through Transform 1 we get:

$$(W_{1} + n_1)^2 + W_2^2 + W_3^2 = r_1^2 \tag {Sphere1 of W}$$ $$(W_{1} - n_2)^2 + W_2^2 + W_3^2 = r_2^2 \tag {Sphere2 of W}$$ $$W_{1}^2 + (W_2 - n_3)^2 + W_3^2 = r_3^2 \tag {Sphere3 of W}$$

Note that we know $n_3 = |C_3 - \Omega|$ because of how $E_2$ was chosen to be defined. By choice of transform and values of $n$, Sphere 1 and Sphere 2 has the solution: $$W_1 = 0$$ $$W_2^2 + W_2^2 = r_x^2$$

Which solves the sphere-sphere intersection part. Note that $r_x^2$ can be negative, which is a complex circle.


Now we are left with solving a circle circle intersection. This is really a simpler case of the same technique from above.

$$W_2^2 + W_2^2 = r_x^2 \tag {Circle 1}$$ $$(W_2 - n_3)^2 + W_3^2 = r_3^2 \tag {Circle2 was Sphere3}$$

Chose transform $U = W - \begin{bmatrix} 0 & m_1 & 0\end{bmatrix}$.

$$(U_2 + m_1)^2 + U_3^2 = r_x^2 \tag {Circle 1 in U}$$ $$(U_2 - m_2)^2 + U_3^2 = r_3^2 \tag {Circle2 in U}$$

If we choose :

$$m_1 + m_2 = m = n_3$$ $$r_x^2 - m_1 = r_3^2 - m_2^2 = r_F^2$$

These conditions solve to :

$$m_1 = \frac {m^2 - r_X^2 + r_3^2} {2 m} $$ $$m_2 = \frac {m^2 + r_X^2 - r_3^2} {2 m} $$ $$r_F^2 = - \frac {\left(m^2 - (r_3 - r_X)^2\right)\left(m^2 - (r_3 + r_X)^2\right)} {4 m^2} $$

Then Circle 1 and Circle 2 solve to : $$U_2 = 0$$ $$U_3^2 = r_F^2$$

Note that $r_F^2$ can be complex in the case of a non-overlapping system.


Going from $U$ to $V$, just backsubstitute the expressions:

$$ U = \begin{bmatrix} 0 & 0 & \sqrt{r_F^2}\end{bmatrix} \tag {This has 2 complex values}$$ $$ W = U + \begin{bmatrix} 0 & m_1 & 0 \end{bmatrix}$$ $$V = W \begin{bmatrix} E_1 \\ E_2 \\ E_3 \end{bmatrix} + \Omega \tag{Transform 1}$$

Applying the formulas for intermediate variables along the way. It gets an analytical solution, but not a short or pretty one.

$\endgroup$
2
  • $\begingroup$ I think this system of equations probably has 3 (not necessarily independent) discriminates, and that the $r_x^2 = $ equation is one of them, and that the other 2 are the cyclic versions of that. There is probably a better solution than this which treats the variables fairly, and uses the 3 discriminates to find a better analytic solution. I'd <i>really</i> be interested if someone could present a solution that treats the variables fairly, rather than picking a few to solve for first. $\endgroup$
    – DanielV
    Nov 12, 2013 at 10:17
  • $\begingroup$ Thanks.It was nicely done. But can you solve this for step by step if we have values for the constants. as an example $S_{1}=(x-3)^2+(y-5)^2+(z-9)^2=119$, $S_{2}=(x+8)^2+(y+11)^2+(z-5)^2=49$, $S_{3}=(x-1)^2+(y+2)^2+(z+3)^2=78$ $\endgroup$
    – Harry
    Nov 13, 2013 at 6:40
1
$\begingroup$

To get you started:

  1. $(x-x_{1})^2+(y-y_{1})^2+(z-z_{1})^2=a^2$
  2. $(x-x_{2})^2+(y-y_{2})^2+(z-z_{2})^2=b^2$
  3. $(x-x_{3})^2+(y-y_{3})^2+(z-z_{3})^2=c^2$

Use 1.+2.

  1. $x^2-2xx_1+x_1^2+y^2-2yy_1+y_1^1+z_1^2-2zz_1+z_1^2=a^2$
  2. $x^2-2xx_2+x_2^2+y^2-2yy_2+y_2^2+z_2^2-2zz_2+z_2^2=b^2$

Now 1. - 2.

$$ -2xx_1+2xx_2+x_1^2-x_2^2-2yy_1+2yy_2+2zz_1-2zz_2+y_1^2-y_2^2+z_1^2-z_2^2=a^2-b^2$$

Simplify

$$2x(x_2-x_1)+2y(y_2-y_1)+2z(z_2-z_1)+(x_1^2-x_2^2+y_1^2-y_2^2+z_1^2-z_2^2)=a^2-b^2$$

$$\Leftrightarrow x=\dfrac{a^2-b^2-2y(y_2-y_1)-2z(z_2-z_1)-(x_1^2-x_2^2+y_1^2-y_2^2+z_1^2-z_2^2)}{2(x_2-x_1)}$$

Now put $x$ into 1. or 2. and solve the quadratic for $y$. Use at last Ari's hint.

$\endgroup$
5
  • $\begingroup$ when 1.+2 what has happened to $z$ terms? $\endgroup$
    – Harry
    Nov 11, 2013 at 9:59
  • $\begingroup$ how do i find $z$? $\endgroup$
    – Harry
    Nov 11, 2013 at 10:02
  • $\begingroup$ Oh sorry, I mixed sphere + circle. Give me a second. $\endgroup$
    – ulead86
    Nov 11, 2013 at 10:21
  • $\begingroup$ when we put $x$ into 1. or 2 and quadratic for $y$ comes with $z$? $\endgroup$
    – Harry
    Nov 11, 2013 at 10:35
  • $\begingroup$ I think this isn't correct. I believe that subtracting (1) and (2) is not a solution preserving operation. Note how you get only 1 value for $x$, but you should get 2 possible complex values for $x$ except in the case where the centers are shared. If this approach was sound then you could also get a single value for $y$ and $z$ the same way you did for $x$, and express the solution as the sum, product, and difference of centers and radii without using square root. $\endgroup$
    – DanielV
    Nov 12, 2013 at 10:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.