Intersecting Circles. Let $O_1$ , $O_2$ be the centers of circles $C_1$ , $C_2$ in a plane respectively, and the circles meet at two distinct points $A$ , $B$ . Line $O_1$$A$ meets the circle $C_1$ at point $P_1$ , and line $O_2$$A$ meets the circle $C_2$ at point $P_2$. Determine the maximum number of points lying in a circle among these 6 points $A$, $B$, $O_1$ , $O_2$ , $P_1$ and $P_2$.
I drew the two circles, but both the points $P_1$ and $P_2$ coincide with the point $A$.   Is this method of solving the question correct?   
 A: Analysis & Solution
Notice that $\angle ABP_1+\angle ABP_2=90^o+90^o=180^o$. Hence, $P_1, B, P_2$ are collinear. Thus, we have three groups of collinear points, which are $(A, O_1, P_1), (A, O_2, P_2)$ and $(P_1, B,P_2).$
On one hand，we may claim that, no matter what $5$ points we pick among the given $6$ points, or we pick all of them, there necessarily exists at least one group of collinear points, which is not concylic. For this reason, we obtain the fact,there exists no more than $4$ concyclic points  among the given $6$ points.
One the other hand, we may pick $4$ points such that they are concylic, for example, we want $O_1, O_2, P_1, P_2$ are concylic. For this purpose, we only need $c_1$ and $c_2$ are equal circles. (Can you prove this?)
Now, we may conclude, the maximum number of points lying in a circle among the six is $4.$

A: If the correction as described in the comments is taken to be true, then you can prove that $P_1, B$ and $P_2$ are collinear. (Sketch it out and first prove that $ABP_1$ is a right angle.)
After proving this, you can reason that a maximum of four of the six points can lie on a circle.
