Prove that three points are enough to draw/define one and only one circle

The set of points that are equidistant from two points $A,B$ is a straight line, right? Precisely, the line that goes through the midpoint of the segment $AB$ and is perpendicular to it. Let's call this line $l(A,B)$.

Given 3 non-collinear points $A,B,C$, the lines $l(A,B)$ and $l(A,C)$ are not parallel, because the lines $AB$, $AC$ are not parallel, and therefore meet in exactly one point. This point is equidistant from $A,B,C$, and is therefore the only such point. It is the center of the unique circle that goes through these three points.

An expansion on the other answer.

More analytically (or explicitly), let $a$,$b$,$c$ be three points. We want to show there is only one point equidistant to all three.

Let $a = (a_1,a_2)$, $b = (b_1,b_2)$,$c = (c_1,c_2)$. Points $x = (x_1,x_2)$ equidistant to all three must satisfy,

$$ (x_1-a_1)^2+(x_2-a_2)^2=(x_1-b_1)^2+(x_2-b_2)^2=(x_1-c_1)^2+(x_2-c_2)^2$$

Which is the system of equations

$$ (x_1-a_1)^2+(x_2-a_2)^2=(x_1-b_1)^2+(x_2-b_2)^2$$ $$(x_1-b_1)^2+(x_2-b_2)^2=(x_1-c_1)^2+(x_2-c_2)^2$$

The solution to each equation is a line. Specifically, the solution to the first equation is the line

$$ (x_1-a_1)^2+(x_2-a_2)^2=(x_1-b_1)^2+(x_2-b_2)^2$$

$$ -2a_1x_1+a_1^2 -2a_2x_2+a_2^2=-2b_1x_1+b_1^2 -2b_2x_2+b_2^2$$

$$ x_1 = \frac{(-2b_2+2a_2)x_2+(b_2^2-a_2^2)}{-2a_1+2a_2}$$

Since two lines intersect at at most one place, the solution to the system is either a single point which defines the center of the circle, or there is no solution (the case of $a$,$b$,$c$ collinear). If we include the point at infinity as Qiaochu mentions then two lines always intersect at exactly one point (with parallel lines intersecting at infinity), and the circle will be through $a$,$b$,$c$ with infinite radius.

Otherwise the radius will be $\sqrt{(x_1-a_1)^2+(x_2-a_2)^2}$ and we have defined a circle.