Can we prove that circle has at most two colinear points? Ok so I stumbled across a problem(I found a solution just to be clear) and it got me thinking.The problem is a classic,it was challenge to prove that line can intersect circle in at most 2 points.
So I found a solution with quadratics,and parametric line and all that but what I thought was better solution is following.
If we prove that circle has a corresponding colinear point for each point that is part of the circle(aka that it is made of pairs of colinear points,which really is true).If we could prove that then we could easily prove that there is no line that intersects the circle at more than two points.
My idea is to prove it by saying that if we take one axis of symmetry for circle then for each point on one side,there is point colinear to it on the other side of the axis,but how can we prove that the number of these colinear points does not exceed two,or does such thing even need to be proven?
 A: Circle is just a set of points that are equidistant to its centre. Say there is a circle and a line that intersects at more than two different points, and let any distinct three of these intersections be $P$, $Q$ and $R$. Without loss of generality, let $Q$ is between $P$ and $R$ on the straight line. Let the centre of the circle be $O$.
$$\begin{align*}
\angle PQO + \angle OQR =& 180^\circ&\text{(angles on a straight line)}\\
\angle PQO =& \angle OPQ&\text{(base angle, }OP=OQ\text{)}\\
\angle OQR =& \angle QRO&\text{(base angle, }OQ=OR\text{)}\\
(\angle OPQ + \angle PQO) + (\angle OQR + \angle QRO) =& 360^\circ\\
(180^\circ-\angle POQ)+(180^\circ - \angle QOR) =&360^\circ &\text{(Internal angles of triangle)}\\
\angle POQ + \angle QOR =&0^\circ
\end{align*}$$
This means either: both angles $\angle POQ$ and $\angle QOR$ are $0^\circ$, or one of these angles is positive and the other triangle has their bases overlapped. In any case, some of the intersection points are shown to be the same.
A: Suppose a circle contains three distinct colinear points $A$, $B$, $C$ (I always write "colinear" without the extra "l" because I don't think it belongs there) where $B$ is beween $A$ and $C$.  Let $L$ be the line through the three points.  Let $O$ be the center of the circle.  Let $L_1$ be the set of all points equidistant from $A$ and $B$.  $L_1$ contains $O$ and is perpendicular to $L$. Let $L_2$ be the set of all points equidistant from $B$ and $C$.  $L_2$ contains $O$ and is perpendicular to $L$.  $L_1$ and $L_2$ intersect at $O$.  Since $L_1$ and $L_2$ are perpendicular to $L$,  $L_1$ and $L_2$ are parallel.  $L_1$ and $L_2$ intersect $L$ at two different points (one of them is halfway between $A$ and $B$ and the other is halfway between $B$ and $C$) so $L_1$ and $L_2$ are disjoint.  This is a contradiction.
A: We can use that the perpendicular bisector of a chord goes through the center of the circle. If there were three points on the same chord, the bisectors would be parallel and two of them would miss the center,
A: A disc is a convex geometrical figure. Any figure with a property that line intersects its boundary at more than two points has to be non-convex. (Just to clarify, I do not count the lines which have no points in the interior of the figure as "intersecting the boundary")
It is because if you go along the line you first enter the figure and then leave the figure and if after this point you enter the figure again, the figure isnt convex. (You can find two points belonging to the figure that if you connect them with a line segment there is going to be a point on that line segment which does not belong to the figure.)
It has been pointed out that the same theorem is not true for a square. What distinguishes a square from a circle is not convexity but the fact that if you consider lines that do not enter the respective figures then in case of a square you can get infinitely many common points but in the case of a circle you can get at most 1 point.
One advantage of this approach is that you can immediately show that there are no 3 collinear points on an ellipse, for example.
A: There have been many fine answers given. However, it might be also nice to mention a more general result: Benzout's Theorem. (See Benzout's Theorem or Benzout's Theorem)
This essentially says an algebraic curve of degree $m$ and an algebraic curve of degree $n$ meet at $mn$ points in the projective plane. So in the real plane, it says they meet at $p \leq mn$ total points.
A circle is an algebraic curve of degree $2$ as a circle has the form $(x-x_0)^2+(y-y_0)^2-r^2=0$ and a line is an algebraic curve of degree $1$ as a line has the form $ax+by-c=0$. 
The number of points $p$ in a circle on the same line, say $ax+by+c=0$, is the same as saying the line and the circle have $p$ intersection points. By Benzout's Theorem, the amount of intersections $p$ must obey $mn=2(1)=2\geq p$. 
So at most two points on a circle can be colinear. You get $2$ intersections if you choose any two distinct points on the circle and $1$ if the points are the same (this is the tangent line to the circle at that point). Since the points forming the line are chosen from the circle, $0$ intersections is of course not an option.  
