3 points and half-sphere

Question

Consider 3 points above the ground. Those clearly define a unique half-sphere which is based on the ground and running through those.

In fact, the construction is obvous: simply consider the plane, defined by the points, construct the circumcenter of the points within that plane, errect the normal there, and you'd get the center of the searched for sphere where it hits the ground.

What I'm looking for so, is a clear function, which does provide the radius of that sphere directly from the coordinates of those points.

--- rk

Answer 1

HINT:

Find circum-center $P$ of the three given points $(A,B,C)$ with given $(x,y,z) $ coordinates. Find unit normal to the plane (shift position vector to $P$ found using cross vector product $ AB \times AC)$. Intersection of normal vector to ground plane is the center of sphere $O$. And radius required is distance of any of three points to $O$.

Answer 2

Suppose three points $(a,b,c),(a',b',c'),(a'',b'',c'')$ are those that problem gives us. Also $(x,y,0)$ the center of sphere on $XoY$ plane, so we have: $$(a-x)^2+(b-y)^2+c^2=(a'-x)^2+(b'-y)^2+c'^2=(a''-x)^2+(b''-y)^2+c''^2=r^2$$ Since $x^2,y^2$ will be clear from all equations, they are linear and easily solve: $$a^2-2ax+b^2-2by+c^2=a'^2-2a'x+b'^2-2b'y+c'^2=a''^2-2a''x+b''^2-2b''y+c''^2$$ Then replace $x,y$ in any of above to get $r^2$.