Equation (with polar coordinates) of circle on a sphere

[mario991]

hi,
i'm a newbie...
i have this problem:
i have a sphere with known and constant R (obvious),
i have two point with spherical coordinates
P1=(R,p_1,t_1) and P0=(R, p_0, t_0)
p_x = phi x = latitude x
t_x = theta x =longitude x
the distance between point is
D= R*arccos[cos(p_0)*cos(p_1)+sin(p_0)*sin(p_1)*cos(t_1-t_0)]
source (http://mathforum.org/library/drmath/view/51882.html)
or similar (http://www.movable-type.co.uk/scripts/latlong.html)

but in spherical coordinates which is explicit formula of the
circle on sphere with center in P0 and radius P0 to P1 ?

many thanks
mario

PS image on this link
https://www.dropbox.com/s/807dqx0wovvw34v/2015-03-24_213647.png?dl=0

PSS i cannot understand if this is the same problem ...
https://www.physicsforums.com/threa...ven-two-points-on-circle.571535/#post-3732362

 

[Simon Bridge]

I take it P0 and P1 are both on a sphere radius R centered at the origin?
You want the spherical-polar coordinates for a circle center P0 that passes through P1?

There are a number of ways to construct it - i.e. you can rotate the axes to that the z axis goes through P0, work out the formula, then rotate back.
What have you tried?

 

[mario991]

I take it P0 and P1 are both on a sphere radius R centered at the origin?
yes
You want the spherical-polar coordinates for a circle center P0 that passes through P1?
yes

There are a number of ways to construct it - i.e. you can rotate the axes to that the z axis goes through P0, work out the formula, then rotate back.
What have you tried?

i have tried many way (intersection plane A x + B y + C z = D with sphere x^2 + y^2 + z^2 = R^2, intersect cylinder (or cone) with sphere) but I'm searching the simpliest algebrically solution ...

PS soultion with programs like Wolframalpha or Derive for R*arccos[cos(p_0)*cos(p_1)+sin(p_0)*sin(p_1)*cos(t_1-t_0)] - D = 0 are to complex and give not the correct solution

 

[Simon Bridge]

How about this:
If P0 is at $\vec r_0$ and P1 is at $\vec r_1$ then these two vectors form a plane.
The circle you want is in the plane perpendicular to $\vec r_0$ that contains the point P1.
The radius of the circle in that plane is given by Pythagoras ... sketch out the vectors and you should see what I mean.

 

[mario991]

Solution found!

,
all on http://demonstrations.wolfram.com/ParametricEquationOfACircleIn3D/
thanks Mr. Simon !

 

[Simon Bridge]

Technically that is not the equation you asked for, but you can certainly use it to find the one you asked for.