Does a colouring of a graph on two colours always have certain kinda of circle Is there a planar set of points $P$   $(|P|\geq 4)$ such that no matter how you colour the points with two colours  you can always find four points on a circle so that all four of the point have the same colour and their circle contains no other points from $P$
I'm not really sure how to start working on this graph problem any advice?
 A: Yes, there is. 
Take 21 points arranged in three horizontal rows and seven vertical columns. For example take the set $P=\{1,2,...,7\}\times\{1,2,3\}$, that is, $P=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),..,(7,1),(7,2),(7,3)\}$

Clearly there are many circles that go through exactly four points of $P$ (which would be the vertices of rectangles with sides parallel to the coordinate axes). I also believe that there is no circle that goes through five or more points of $P$. I would need the latter property, but since I do not want to verify it, I would just circumvent this detail by moving some of the columns a tiny little bit, to the left or to the right (keeping them vertical), if necessary so as to ensure that no circle goes through five or more points of $P$. (Edit. It is indeed necessary to shift columns (and/or rows) a bit to ensure no circle goes through five or more points of $P$. Otherwise, it is clear from the picture that there is a circle centered at $(2.5,1.5)$ that goes through six points of $P$.) 
So now color the points of $P$ two colors. There could be at most two points at the bottom row (on the line $y=1$), each with the property that the point just above it (on the line $y=2$) has the same color. Indeed, if there were three points on the bottom row with this property, then two of them would have to have the same color, and these two points together with the two points just above them would be the vertices of a rectangle (with height 1), and they would determine a monochromatic circle. 
So, discarding at most two columns, we would have at least five columns left with the property that each point at the bottom row ($y=1$) is colored differently than the point just above it ($y=2$). A similar argument shows that among these, there are at most two points at the bottom row ($y=1$), each with the property that the point above it on the top row ($y=3$) has the same color (as the point at the bottom). So discarding at most two more columns, we are left with at least three columns with the property that for each point at the bottom, each of the two points above it (on rows two and three) is colored with the opposite color (compared to the point at the bottom). This means that in each of these (three or more) columns under consideration, the points on the second and the third row share a color. But since we have at least three columns remaining, there must be two points on row two ($y=2$) that share a color: These two points together with the points above them on row three form a rectangle with monochromatic vertices, and determine a circle with four monochromatic points in $P$. 
