First, you may wish to demonstrate that the basic premise is false.
Try (0,0,0), (1,0,0), and (2,0,0)
Write a new definition of the problem and maybe there's a way to proceed.
for any 3 points in 3d space. (x1,y1,z1) etc there is a unique circle that can be drawn throught then. I have spent many hours trying to derive the equasion of that circle so that I can then find the intersection of the plane Y=a and that circle but have not got anything clean, the approach that I have tried is to translate and rotate the coordinate system so that the 3 points all lie on the plane z=0 then use a reverse transformation to get the actual intersection, this is horendous does anyone have a better approach, any assistance would be of help I'm stumped
Denote by v_1, v_2 and v_3 the position vectors of the three points. Then will be a vector in the direction normal to the plane of the circle. Form this cross product, and divide by its length to get a unit vector w normal to the circle.
Next, notice that for any real number λ, the vector represents a point on the perpendicular bisector of the line joining v_1 and v_2 (in the plane of the circle). Similarly, for any μ, is a point on the perpendicular bisector of the line joining v_1 and v_3 (in the plane of the circle). You can solve the equation finding λ and μ by comparing the three coordinates of the vectors in that equation. The vector is the centre of the circle. (The equation will have a solution whenever the three vectors v_1, v_2 and v_3 are non-collinear.)
Then is a vector from the centre to a point on the circle, and is a perpendicular vector from the centre to a point on the circle. So the equation of the circle is given parametrically as the locus of the vector .
If you want to know where the circle cuts the plane y=a, all you have to do is to set the y-coordinate of p equal to a, and solve for θ.
Let be three distinct non-collinear points.
There is a circle, ( ) passing through all three points.
By substituting the points (here are arbitrary point on circle) we get:
This is a homogenous system for , it has non-trivial solutions.
Evaluating the determinant will give you the equation of the circle.
Note, this method generalizes to all conics: lines, parabolas, ellipses,....
But computing that determinant is a nightmare, it gets even more horrific with higher order conics, so unless you got a computer program stay away.
PerfectHacker's solution is very elegant in 2 dimensions. It is easily extended to 3 dimensions (origional problem).
We first demonstrate that the equation of the 3D circle is of the form:
a(x^2 + y^2 + z^2) + bx + cy + d = 0 - (1)
then perfectHacker's solution is easily extended. There is one more part to this though. We need to use this technique to find the plane containing the points:
| x y z 1 |
| x1 y1 z1 1 |
| x2 y2 z2 1 | = 0 - (2)
| x3 y3 z3 1 |
Thus z can be expressed as a linear function of x and y (being on the plane)
and eliminating z in the circle equation, one gets a projection of that circle in X-Y space (an ellipse). From this, one can solve for y as a function of x (two roots of the quadratic), and use its value in the plane equation to also find z in terms of x.
The only part left for me is to explain how I got the form of the circle in equation (1).
There are an infinite number of spheres containing the 3 points on its surface. (imagine any sphere with a large enough radius..it can be moved so that the 3 points touch it).
Equation for the sphere is (vec X - vec R)^2 = r^2 where vec R is the center of the circle, and vec X is the locus of all points on its surface. This is of the form:
x^2 + y+2 + z^2 + gx + hy + iz + j = 0
Since the points lie on a plane, one can expres z as a linear function of x and y, thus obtaining the form as in equation (1)