The answer is $8$

I just could construct an example and nothing more very interesting problem!

I have a very long solution And this problem is an old result First let fn be the max number of unit circle Of n points then it's known that f5=4 f6=8 f7=12 I will prove f5=4 and f6=8 First prove f5=4(then its easy to get f6=8) 1)if 4points are the Orthcenter and vertex of a triangle Then not more than 4 2)if 4points ABCDare on a unitcircle then no more Than5,if there5unitcircle it must be ABCD,ABE,BCE,CDE,DAE Then we get ABCD is a parallelogram E must be the orthcenter of some triangle Condradiction. 3)then if more than 4 unitcircles it must Be ABC BCD CDE DEA EAB then discuss the convex hull of the five points and can have condradiction by chasing angle

And for f6 is the same The example is ABCDdiamond EFonAC AE=CF ABCE unitcircle

duanby wrote: And for f6 is the same The example is ABCDdiamond EFonAC AE=CF ABCE unitcircle Can you post the details of your solution on Chinese forum ? Thanks

For f6 if the numbers of the unit circles more than 8 then no four points on a unit circle and no orthcenter and vertex of triangle the only case is that ABF,BCF,CDF,DEF,AEF,ABD,AEC,BCE,DEBare unit circles but for BCDEF the numbers of unit circles is 5 contraction.

duanby wrote: And for f6 is the same The example is ABCDdiamond EFonAC AE=CF ABCE unitcircle Could anyone please elaborate this?