Is this curve the circumference of a circle? Let $\Gamma$ be a single closed curve with no self-intersections on a plane which satisfies the following condition : 
Condition : For any distinct four points $P, Q, R, S$ on $\Gamma$, if the line $PQ$ is orthogonal to the line $RS$, then these four points lie on a circle.
Question : Is $\Gamma$ the circumference of a circle?
Motivation : I like this kind of question such as this question which got me interested in the above question. The answer seems yes, but I'm facing difficulty. Can anyone help?
 A: This is a start; the remainder would be explicit formulas in coordinates.
Your curve is compact, there is a pair of points $P,Q \in \Gamma$ whose distance realizes the diameter of the set. Take the perpendicular bisector of the segment $PQ.$ this meets $\Gamma$ in two new points $R,S.$ These lie on a circle, of which $RS$ is a diameter. Now, if you draw a sample segment $PQ,$ and draw a number of circles passing thorough those two points, you will find that the circle of smallest diameter is that for which $PQ$ is already a diameter. It follows that this is the precise circle on which $R,S$ lie. So, without any loss of generality, we may place points of $\Gamma$ on $x,y$-axes at $$ P=(0,1), Q = (0,-1), R = (1,0), S = (-1,0). $$ It also follows that no point of $\Gamma$ can lie outside the square $$ -1 \leq x \leq 1, \; -1 \leq y \leq 1 $$ as it would then be of distance greater than $2$ from one of $P,Q,R,S.$
Finally, given any point $T =(a,b) \in \Gamma,$ say with $a > 0, b > 0$ your four-point rule says that we can find its explicit "reflection" across the $x$-axis by writing down the circle through $T,R,S$ and  computing $T_x = (a,b').$ We may then find that reflection across the $y$-axis, call it $T_{xy} = (a', b').$ Or, we could begin by reflecting across the $y$ axis, with $T_y = (a'', b)$ followed by $T_{yx} = (a'', b'').$ My suspicion is that, if $a^2 + b^2 \neq 1,$ the distance between either $TT_{xy}$ or $TT_{yx}$ exceeds $2,$ which is prohibited as that is what we are now calling the distance between $PQ.$ Or, one of $T_x, T_y,T_{xy}, T_{yx} $ lies outside the standard  square indicated.
EDIT: with  $T =(a,b),$ I get
$$  T_x = (a, \frac{a^2 -1}{b}),  $$  and 
 $$  T_y = ( \frac{b^2 -1}{a},b).  $$ So, if $a^2 + b^2 = 1,$ both $T_x$ and $ T_y$ also lie on the unit circle. 
EEEDDDIIIIITTTT: Probably necessary to shrink the square to the overlap of the closed disks around each of $P,Q,R,S$ of radius $2.$ This shape has four corners, one is at $(t,t)$ where $t = \frac{\sqrt 7 - 1}{2} \approx 0.823$
EEEEEEEDDDDDDDDDIIIIIIITTTTTT: At this pont, probably the most efficient thing is to find the point $T$ on $\Gamma$ farthest from the origin; by reflecting or rotating the square, we can require this to be $(x,y)$ with $0 < x,y < 1$ and $x^2 + y^2 > 1;$ this last piece is the assumption. I suspect either $T_{xy}$ or $T_{yx}$ will be farther from the origin than $T,$ contradicting assumption. And that will complete the proof. I worked it out, it does not appear that either  $T_{xy}$ or $T_{yx}$ will necessarily be farther from the origin than $T.$ Not a bad idea, though...Note that $P,Q,R,S$ are distance $1$ from the origin; one can play with that.
http://en.wikipedia.org/wiki/The_Rule_of_Four
