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One sphere with radius $R$, named big sphere, two points on it: $a (\text{(longitude)}_a, \text{(latitue)}_a), b(\text{(longitude)}_b, \text{(latitude)}_b)$, $\text{distance}\space (a,b)=r$, $a$ is center & $r$ is radius. There is another sphere, named little sphere. Now what is the equation of circle of the intersection of two spheres? Parametric equation of the circle: $\text{longitude}=f(xx), \text{latitude}=g(xx)$

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assume the big sphere equation is:x^2+y^2+z^2=R^2, then the little sphere is:(x-xa)^2+(y-ya)^2+(z-za)^2=r^2,but i want the parameter equation: lon=f(x),lat=g(x) 0<=x<2pi – user5900 Jan 16 '11 at 8:35

3 Answers 3

This (calculating the equation of the circle intersection between two spheres) will probably help.

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I assume $\text{distance}\ r$ is an Euclidean distance in 3D, not the length of the great circle's arc on the big sphere. Of course $0 \le r \le 2R$, with each 'equal' case causing a circle to degenerate to a single point.

Let's define a Cartesian coordinates system, so that:

  • the origin is in the big sphere's center,
  • XY plane is a plane of equator: $\varphi = 0$,
  • positive latitudes $\varphi$ correspond to positive $z$,
  • zero longitude $\lambda = 0$ corresponds to XZ plane $y=0$,
  • longitude is measured from positive X semiaxis towards positive Y semiaxis.

Then: $$ \begin{cases} x = R \cos \varphi \cos \lambda \\ y = R \cos \varphi \sin \lambda \\ z = R \sin \varphi \end{cases} \quad \text{and}\quad \begin{cases} \varphi = \arcsin \frac zR \\ \lambda = \arctan \frac yx \end{cases} $$ Of course one needs to check for singularity when calculating $\lambda$ for $x=0$ and adjust the result by $+180^\circ$ when $x<0$.

Let $\rho$ = an angle between the big sphere's radii ending at $a$ and at $b$. Then $\rho = 2\arcsin \frac r{2R}$.

Suppose $a = a_1$ = 'the North Pole': $\varphi_{a1} = 90^\circ, x=y=0, z=R$. Then the circle sought $C$ is a circle of latitude at $\varphi = 90^\circ - \rho$ (with positive latitudes corresponding to the 'northern' hemisphere).
It can be easily parameterized as

$$\begin{cases}\varphi_1 = 90^\circ - \rho \\ \lambda_1 = t\end{cases} \ \ \text{for}\ \ 0<t<360^\circ$$

and then transformed into XYZ coordinates:

$$ \begin{cases} x_1 = R \cos \varphi_1 \cos t \\ y_1 = R \cos \varphi_1 \sin t \\ z_1 = R \sin \varphi_1 \end{cases} $$

Now suppose we shift $a$ along the Greenwich meridian to the south, to some point $a_2$ at latitude $\varphi_a$. That means the whole circle $C$ undergoes a rotation in 3D about the Y axis by an angle $\delta_\varphi = 90^\circ-\varphi_a$, measured from positive Z semiaxis towards positive X semiaxis. Each point of circle $C$ gets its coordinates tranformed like this:

$$\begin{cases} x_2 = x_1 \cos \delta_\varphi + z_1 \sin \delta_\varphi \\ y_2 = y_1 \\ z_2 = z_1 \cos \delta_\varphi - x_1 \sin \delta_\varphi \\ \end{cases}$$

Now we need to shift $C$ along the circles of latitude so its 'center' gets into a desired position $a_3$ at longitude $\lambda_a$. This is a 3D rotation about the Z axis:

$$ \begin{cases} x_3 = x_2 \cos \lambda_a - y_2 \sin\lambda_a \\ y_3 = y_2 \cos \lambda_a + x_2 \sin \lambda_a \\ z_3 = z_2 \end{cases} $$

Now you only need to compose the transformations described above and possibly add another $(x,y,z)\to(\varphi,\lambda)$ transformation to get a parametric equation of the circle.

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The circle belongs to a plane that can be found by subtracting the equations of both spheres:


Now plug the spherical coordinates to yield


You can express $\theta$ as a function of $\phi$ or conversely, by solving a linear trigonometric equation of the form

$$\alpha\cos t+\beta\sin t=\gamma.$$

WLOG (normalize if necessary), $\alpha=\cos u$ and $\beta=\sin u$, then $\cos(t-u)=\gamma$ gives $t$.

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