Consider 3 points on a plane, points are real. Is it possible that the points are placed in a way that makes it impossible to draw a circle trough them.
- I know that if the point forms a line then its not a circle (unless we allow for a circle with a infinite radius to pass trough the points). But are there any other cases?
- I know there are cases where theres more than one solution, such as when 2 or 3 of the points are the same.
Now i understand that the three point circle forms 2 triangles and therefore I think there does not exist cases other than points being in the line that dont work. But how do i build a proof out of this?
Can i use the fact that a normal to a point in between the points allways has a intersection somewhere if they are not perpendicular. And then just state that this only happens if the points form a line?
As noted in the comments ( and in your add to the question) we have a simple geometric solution of this problem simply noting that the axis of two non aligned consecutive segments always intersect in a point that is the center of the circumference passing through the extremes of the segments . You can find also an analytical solution noting that a circumference passing through three points: $A=(x_A,y_A)$, $B=(x_B,y_B)$, $C=(x_C,y_C)$ has an equation $x^2+y^2+ax+by+c=0$ such that: $$ \begin{cases} ax_A+by_A+c=-x_A^2-y_A^2\\ ax_B+by_B+c=-x_B^2-y_B^2\\ ax_C+by_C+c=-x_C^2-y_C^2 \end{cases} $$ that is a system in the unknowns $a,b,c$. We can show that this systems has no solutions if the three point are on the same line and has infinite solutions if at least two points coincide.