In Euclidean and projective geometry, five points determine a conic (a degree-2 plane curve), just as two (distinct) points determine a line (a degree-1 plane curve). There are additional subtleties for conics that do not exist for lines, and thus the statement and its proof for conics are both more technical than for lines.
Formally, given any five points in the plane in general linear position, meaning no three collinear, there is a unique conic passing through them, which will be non-degenerate; this is true over both the Euclidean plane and any pappian projective plane. Indeed, given any five points there is a conic passing through them, but if three of the points are collinear the conic will be degenerate (reducible, because it contains a line), and may not be unique; see further discussion.
This result can be proven numerous different ways; the dimension counting argument is most direct, and generalizes to higher degree, while other proofs are special to conics.
Intuitively, passing through five points in general linear position specifies five independent linear constraints on the (projective) linear space of conics, and hence specifies a unique conic, though this brief statement ignores subtleties.
More precisely, this is seen as follows:
conics correspond to points in the five-dimensional projective space
requiring a conic to pass through a point imposes a linear condition on the coordinates: for a fixed the equation is a linear equation in
by dimension counting, five constraints (that the curve passes through five points) are necessary to specify a conic, as each constraint cuts the dimension of possibilities by 1, and one starts with 5 dimensions;
in 5 dimensions, the intersection of 5 (independent) hyperplanes is a single point (formally, by Bézout's theorem);
general linear position of the points means that the constraints are independent, and thus do specify a unique conic;
the resulting conic is non-degenerate because it is a curve (since it has more than 1 point), and does not contain a line (else it would split as two lines, at least one of which must contain 3 of the 5 points, by the pigeonhole principle), so it is irreducible.