This problem seems to be a bit easy. If any $4$ points are concyclic then clearly $k=0\implies $ any $4$ of the $6$ points are concyclic. So if we fix a $\triangle $ among these $6$ points, then the other $3$ points lie on its circumcircle $\implies$ all $6$ points lie on a circle. So now assume that no $4$ points are concyclic. For any $4$ points, call the point having power $k $ WRT the circle passing through the other three points the pivot of that set of $4$ points. Since there are $\binom {6}{4}=15 $ ways of choosing $4$ points there are atleast $15$ pivots. But there are only $6$ points, so there is a point acting as a pivot for three sets of $4$ points. Let this point be $A $ and the other $5$ points constitute a set $T=\{A_i\forall 1\leq i\leq 5\}$. Let the three circles for which $A $ is a pivot be $X,Y,Z $ and the points among $T $ through which they pass form the sets $P,Q,R $ respectively $\implies |P|=|Q|=|R|=3$ and $A $ is the radical centre of $X,Y,Z \implies |P\cap Q\cap R|=0$. Since $P\cup Q\cup R$ is a subset of $T $, it follows that $5\geq |P\cup Q\cup R|=\sum |P|-\sum |P\cap Q |+|P\cap Q\cap R|=9-\sum |P\cap Q|$ so atleast one of $|P\cap Q|, |Q\cap R|, |R\cap P|$ has atleast $2$ elements. WLOG we assume that $A_1,A_2\in P\cap Q\implies A_1A_2$ is the radical axis of $X $ and $Y \implies $ the radical centre $A\in A_1A_2$ which contradicts their non- collinearity. Hence we are done.

WLOG, let no four points be on a circle; else the conclusion is clearly true. Note that each quadruple of points is assigned a point having this special property. Since there are $6$ points and $\binom{6}{4}=15$ quadruples, at least one point is a member of at least $3$ quadruples. It is clear that we cannot choose three $3$-element subsets from a set of $5$ elements such that no two of the $3$-sets have more than $2$ elements in common. Consequently, two of the quadruples have a pair of elements, other than the chosen point, in common. By radical axis theorem, we get that these three points are collinear, contradiction! Am I doing something wrong? This looks too simple for Iran TSTs...

Quote: ...such that no two of the $3$-sets have more than $2$ elements in common.... I think you mean less than $2$...

Solved with minakshee Note that if any four points are concyclic, then $k = 0$ which means all $6$ points are on the same circle. So all the $15$ quadruples of points have a unique circle. To each circle, assign a point which has power $k$ with respect to it. Since there are only $6$ points, some point must have same power to three circles and hence is the radical center, say point $F$. Now, the three such circles must be formed by at least three points in each, so they need $9$ points, with multiplicity, to lie on these circles. But since there are only $5$ points, some four must lie on the intersections of these circles. But then some two must be on the intersections of the same two circles and hence the line joining them goes through $F$, a contradiction. $\blacksquare$

If any four points lie on a circle, then $k=0$ , as desired.Asumme that no four points lie on a circle . Call a point $a$ SPHS of a group $(a,b,c,d)$ if power of $a$ wrt $(bcd)=k$. Now, there are a total of $\binom{6}{4}=15$ 4-tuplets, so PHP gives that there exists a point which is SPHS of at least $3$ groups. Thus the point $a$ us the radical centre of the three circles obtained by deleting $a$ from the three groups.If any two have two points in common from , the set of 6 points , then we will have that $a$ will lie on the common chord formed by those two points , a contradiction. But , we can easily see with some work that this will always be the case.( like if two circle are $(bcd)$ and $(bef)$ , then the third one must share atleast two points with one )So , there must exist $4$ points on a circle , and we are done.

We have $\binom{6}{4} = 15$ for-tuple of points and just $6$ points , so there is a point $A$ that has power $k$ to at least $15/6$ circles so there are $5$ points and $3$ 3-tuple , it should be a pairwise $BC$ that exists in two of these three three-tuple , so powers of $A$ to $BCD,BCE$ are equal , it means $A,B,C$ are collinear ( it is not possible) or $B,C,D,E$ are in same circle , now give these $4$ points , we got $k=0$ and $6$ points are in a same circle.