Find length of intersection between 2 points and a sphere

Question

I have a sphere and 2 points. The points have (x,y,z) coordinates and the sphere is defined by its centre (0,0,0) and radius R. I am trying to find the length between the 2 points which intersects the sphere. How can I obtain the equation to describe this length?

See below, my objective is Length, L:

Answer 1

The sphere is: $ x^2 + y^2 + z^2 =R^2 $ .

The line is:
$ x=x \\ y=x\cdot\frac{y_2-y_1}{x_2-x_1} -\frac{x_1(y_2-y_1)}{x_2-x_1} +y_1 =:r_yx+y_0 \\ z=x\cdot\frac{ z_2-z_1 }{x_2-x_1} -\frac{x_1(z_2-z_1)}{x_2-x_1} +z_1 =:r_zx+z_0$.

Sphere and line intersect when:

$ x^2 + (r_yx+y_0)^2 + (r_zx+z_0)^2 =R^2 \implies \\ x^2\cdot(1+r_y^2+ r_z^2) + x\cdot(2r_yy_0+2r_zz_0)+(y_0^2+z_0^2-R^2)=0\\ \\ $

Solving for $x$ : $$ x=\frac{ -(2r_yy_0+2r_zz_0) \pm \sqrt{(2r_yy_0+2r_zz_0)^2-4(1+r_y^2+ r_z^2)(y_0^2+z_0^2-R^2)}}{2(1+r_y^2+ r_z^2)} $$ So for the intersection: $$ (\Delta x)^2 = \frac{(2r_yy_0+2r_zz_0)^2-4(1+r_y^2+ r_z^2)(y_0^2+z_0^2-R^2)}{(1+r_y^2+ r_z^2)^2} \\ (\Delta y)^2 =(\Delta x)^2(\frac{y_2-y_1}{x_2-x_1})^2\\ (\Delta z)^2 =(\Delta x)^2(\frac{z_2-z_1}{x_2-x_1})^2\\ $$ Length of the line segment is: $$ \Delta x \cdot \sqrt{1 +(\frac{y_2-y_1}{x_2-x_1})^2 + (\frac{z_2-z_1}{x_2-x_1})^2} = \\ \sqrt{ \frac{(2r_yy_0+2r_zz_0)^2-4(1+r_y^2+ r_z^2)(y_0^2+z_0^2-R^2)}{(1+r_y^2+ r_z^2)^2}}\cdot\sqrt{1 +r_y^2 + r_z^2}= $$

$$ 2\cdot\sqrt{ \frac{(r_yy_0+r_zz_0)^2-(1+r_y^2+ r_z^2)(y_0^2+z_0^2-R^2)}{(1+r_y^2+ r_z^2)}} $$

Answer 2

Let's call the given two points (vectors) as $$ P_{\,1} = \left( {x_{\,1} ,y_{\,1} ,z_{\,1} } \right)\quad P_{\,2} = \left( {x_{\,1} ,y_{\,1} ,z_{\,1} } \right) $$ then a generic point $P$ on the line connecting the two points will be given by $$ P \in \overline {P_{\,1} P_{\,2} } \quad \Rightarrow \quad P = t\,P_{\,1} + \left( {1 - t} \right)P_{\,2} $$ and when $0 \leqslant t \leqslant 1$ the point will be internal to segment, and otherwise external.
The squared modulus of $P$ represents the square of its distance from origin and is given by the dot product by itself: $$ \left| P \right|^{\,2} = P \cdot P = t^{\,2} \,P_{\,1} \cdot P_{\,1} + \left( {1 - t} \right)^{\,2} P_{\,2} \cdot P_{\,2} + 2t\left( {1 - t} \right)P_{\,2} \cdot P_{\,1} $$ The various dot products are scalars, and quite simple to calculate, and you have a simple quadratic equation.

From here you can follow two possible approaches:

1. Find the points and then their distance
   The point $P$ is constrained on the line, then to constained it to lie on the sphere, just equate $\left| P \right|^{\,2} $ to $R^2$, and solve for $t$.
   The values obtained will tell you whether the points are internal/external to the segment. The two intersection points will then be $$ P_{a,\,b} = t_{a,\,b} \,P_{\,1} + \left( {1 - t_{a,\,b} } \right)P_{\,2} $$ and distance will follow obviously.
2. Find the distance and then the points
   Since your objective is the length of the segment, then you can more quickly derive the expression of $\left| P \right|^{\,2}$ with respect to $t$ and find the value $t_0$ that minimize it. $$ \begin{gathered} 0 = \frac{d} {{dt}}\left| P \right|^{\,2} = 2P_{\,1} \cdot P_{\,1} \;t - 2\left( {1 - t} \right)P_{\,2} \cdot P_{\,2} + \left( {2 - 4t} \right)P_{\,2} P_{\,1} \hfill \\ \Rightarrow \left( {P_{\,1} \cdot P_{\,1} + P_{\,2} \cdot P_{\,2} - 2P_{\,2} P_{\,1} } \right)\;t - P_{\,2} \cdot P_{\,2} + P_{\,2} P_{\,1} = 0 \hfill \\ \end{gathered} $$ This will correspond to the point on the segment which is closest to the origin. Therefore $$ R^{\,2} - \left| P \right|^{\,2} (t_{\,0} ) = \left( {L/2} \right)^{\,2} $$ and you can bypass solving the quadratic equation.
   Then to find the points of intersection, if you need them, consider that $t$ as defined above is the measure of the relative distance (i.e. "with sign") of the point $P$ from $P1$ along the line, in the direction from $P_1$ to $P_2$, and normalized vs the absolute distance between $P_2$ and $P_1$.
   Therefore $$ P_{a,\,b} = P\left( {t_{\,0} \pm \frac{{L/2}} {{\left| {P_{\,2} - P_{\,1} } \right|}}} \right) $$