It is wellknown that if A,B,C are noncollinear points, then there is a
unique circle through the points. It is reasonable to ask how many points
are required to specify a conic. The best way to tackle this is to work in
projective geometry.
Recall that a projective conic C has an equation of the form
F(x,y,z) = ax^{2}+bxy+cy^{2}+fzx+gyz+hz^{2} = 0.
A point on C gives an equation linear in the coefficients of F. for example,
if U[1,1,1] lies on C, then a+b+c+f+g+h = 0.
Observe that, for a nonzero k, kF defines the same conic as F, so we may
guess that five ppoints should be sufficient to determine a conic. Given
five ppoints, we get five equations in the six coefficients a,b,c,f,g,h. These
are consistent since they have a solution a=b=c=f=g=h=0, though this does
not give a conic. Since there are five equations in six variables, the general
solution will involve at least one parameter. To prove that there is a single
locus, we need to show that there is exactly one parameter, so that the
general solution has the form {k(a,b,c,f,g,h) : k real}.
Of course, we need to check that the solution does not yield a degenerate
locus. Since a degenerate locus with several points consists of at most two
plines, at least three of the five ppoints must lie on one pline. Thus, if
we choose five ppoints with no three collinear, then the resultant locus
will be a (nondegenerate) conic.
Our proof uses the Fundamental Theorem of Projective Geometry to reduce
the problem to the case where four of the ppoints are the standard ppoints
X[1,0,0],Y[0,1,0],Z[0,0,1],U[1,1,1].
The Five Point Theorem
If A,B,C,D,E are ppoints, with no three collinear, then there is a unique
conic through all five ppoints.
proof
The CabriJava applet below illustrates the theorem. You can move any of
the points A,B,C,D,E and see the conic. With a steady hand, you may be able
to drag E onto the segment and see the locus degenerate into a line pair.
This is easier if you make AB hoizontal.

