Finding circle of a sphere through two points

Question

We have two points $P_1, P_2$ on a sphere $S$ of radius $R$. Suppose for $r \ll R$, the distance between $P_1$ and $P_2$ is less than $2r$.

Then, $P_1$ and $P_2$ both lie on exactly two radius-$r$ circles of the sphere. This is clear: of the set of planes that contain $P_1$ and $P_2$, there are two of those planes that intersect $S$ to form a circle with radius $r$.

Given $P_1$, $P_2$, $R$ and $r$, how can I calculate the centers of these two circles?

I would prefer to do as little messy spherical geometry as possible :).

Answer 1

I'm assuming that you want a circle in space, with its center inside $S$ and radius $r$ measured along a straight line in space. If you instead want a circle on the sphere, i.e. with center a point on the sphere, and radius measured along a geodesic arc, this answer doesn't exactly apply. But for $r\ll R$ it might still be useful. Otherwise, projecting the center in space back onto the sphere is easy, so all you'd have to do is translate the geodesic radius to a straight line radius.

The distance between the center of the sphere and the center of one of these circles has to be

$$ d = \sqrt{R^2 - r^2} $$

This can be seen from the following cross section through the sphere:

So you are looking for a point $C$ such that $\lVert P_1-C\rVert=\lVert P_2-C\rVert=r$ and $\lVert C\rVert = d$. If we use coordinates

$$P_1=(x_1,y_1,z_1)\qquad P_2=(x_2,y_2,z_2)\qquad C=(x,y,z)$$

then you can rewrite this to

$$(x_1-x)^2 + (y_1-y)^2 + (z_1-z)^2 = (x_2-x)^2 + (y_2-y)^2 + (z_2-z)^2 = r^2\\ x^2 + y^2 + z^2 = R^2 - r^2$$

At that point you can fire up your computer algebra system to find the solutions here. If you want to continue manually, rewrite the equatoons like this:

\begin{align*} x_i^2 - 2x_ix + x^2 + y_i^2 - 2y_iy + y^2 + z_i^2 - 2z_iz + z^2 &= r^2 \\ r^2 + 2x_ix + 2y_iy + 2z_iz - x_i^2 - y_i^2 - z_i^2 &= x^2 + y^2 + z^2 \\ r^2 + 2x_ix + 2y_iy + 2z_iz - R^2 &= R^2 - r^2 \\ x_ix + y_iy + z_iz &= R^2 - r^2 \end{align*}

So at that point you have two linear equations:

$$\begin{pmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \end{pmatrix}\cdot C=\begin{pmatrix} R^2 - r^2 \\ R^2 - r^2 \end{pmatrix}$$

Since you have two equations for three variables, you get a one-dimensional space of solutions: $C\in\{v_1+\lambda v_2\mid\lambda\in\mathbb R\}$. Now revisit the last equation:

$$x^2+y^2+z^2 = \lVert C\rVert^2 = \langle v_1+\lambda v_2,v_1+\lambda v_2\rangle = \langle v_1,v_1\rangle + 2\lambda\langle v_1,v_2\rangle + \lambda^2\langle v_2,v_2\rangle\\= \lVert v_1\rVert^2 + 2\lambda\langle v_1,v_2\rangle + \lambda^2\lVert v_2\rVert^2 = d^2 = R^2 - r^2$$

So that's a quadratic equation in $\lambda$, and its two solutions describe your two circle centers.

Answer 2

Here is a partial answer, if you don't mind a bit of linear algebra.

Let $u_1$ and $u_2$ be the vectors from the center of the sphere to $P_1$ and $P_2$ respectively. These vectors have length $R$. The points you are looking for will each correspond to vectors of length $R$ as well; let us call either of them $x$.

So you want a vector $x$ of length $R$ so that $$u_i\cdot x=R^2\cos\theta$$ for $i=1,2$. Here $\theta$ is the angle of the cone defined by the center of the spehere and the circle of radius $r$. (The exact value depends on whether $r$ is measured along the surface of the sphere or in the plane spanned by the circle.)

Writing $x=x_1u_1+x_2u_2+x_3u_3$, where $u_3$ is a suitable vector orthogonal to $u_1$ and $u_2$ (for example their cross product), you get two linear equations from the above equation. These involve only $x_1$ and $x_2$, so you get a unique solution. Then $x_3$ can be found from the requirement that the length of $x$ be $1$. There will be two solutions to this, a simple quadratic equation.

While carrying out the detailed calculations, don't forget that $u_1$ and $u_2$ are not orthogonal.

Answer 3

Let $M$ be the midpoint of $P_1 P_2$ and $O_1,O_2$ the centres of the circles of radius $r$ through $P_1$ and $P_2$. Let $O$ be the centre of the sphere and $N$ the intersection between the ray $OM$ and the sphere. Let $2x$ be the distance between $P_1$ and $P_2$. By the Pythagorean theorem, we have: $$ O_1 M = O_2 M = \sqrt{r^2-x^2},\qquad OM=\sqrt{R^2-x^2}$$ and since $OO_1=R$, we know the side lengths of $OO_1 M$ and $OO_2 M$. Using the cosine theorem is now easy to find the distance between $O_1$ and the $OM$-line, hence the distance between $O_1$ and $O_2$.