Iranian Mathematical Olympiad

The points should be on a circle (or line) and $A$, $B$ should separate $A^{\prime }$, $B^{\prime }$ on it. To prove necessity, first suppose that the points are not coplanar. Then there exist two parallel planes passing through $A$, $B$ and $A^{\prime }$, $B^{\prime }$ respectively. Any two circles in these planes are not linking. So the points should be coplanar.

Now suppose $B^{\prime }$ is not on the circumcircle of $AB{A}^{\prime }$ (which can be a line). So we can slightly change the circle to find a circle passing through $A$, $B$ such that $A^{\prime }$, $B^{\prime }$ are both outside or both inside it. Now, this circle is not linking with the circle with diameter $A^{\prime }{B}^{\prime }$ orthogonal to the plane containing the points.

So the points should be on a circle (or line). Now, suppose $A$, $B$ do not separate $A^{\prime }$, $B^{\prime }$ on the circle. If we change the circle slightly, still passing through $A$, $B$, then $A^{\prime }$, $B^{\prime }$ will be both inside or both outside the new circle and we arrive to a contradiction like the previous case. So, the necessity of the condition is proved.

To prove sufficiency, Let $C$, $C^{\prime }$ be two different circles passing through $A$, $B$ and $A^{\prime }$, $B^{\prime }$ respectively. Let $P$, $P^{\prime }$ be the planes containing $C$, $C^{\prime }$ respectively. If the points are collinear, then $C^{\prime }\cap P$ is consisted of a point inside $C$ and a point outside $C$. So $C$, $C^{\prime }$ are interlocked. So, suppose the points are on a circle. Let $M$ be the intersection of the segments $AB$ and $A^{\prime }{B}^{\prime }$. We have $MA\cdot MB=M{A}^{\prime }\cdot M{B}^{\prime }$. Let

$l=P\cap P^{\prime }$ which passes through $M$. $M$ is inside $C$, so $l$ intersects $C$ at two points like $X,Y$ and $M$ is between $X,Y$. Similarly, $l$ intersects $C^{\prime }$ at $X^{\prime },Y^{\prime }$ namely, and $M$ is between $X^{\prime },Y^{\prime }$. Suppose $X,X^{\prime }$ are in one side of $M$. We have $$MX\cdot MY=MA\cdot MB=M{A}^{\prime }\cdot M{B}^{\prime }=M{X}^{\prime }\cdot M{Y}^{\prime }$$ So if $MX\le M{X}^{\prime }$, then $MY\ge M{Y}^{\prime }$ and vice versa. So the points of $C^{\prime }\cap P=\{X^{\prime },Y^{\prime }\}$ are in different sides of $C$ or both are on $C$. So $C,C^{\prime }$ are linking and sufficiency of the condition is proved. $\square$