I think it means that from any point on the circle, then 2 tangents to the figure (call it $F$) are perpendicular. Let $A$ be a point on the circle, and $AB,AD$ are tangents to $F$, with $B,D$ on the circle. Then consider the other tangent $BC$ at $B$, with $C$ on the circle. Then we have $ABCD$ is a rectangle and the sides are tangent to $F$. Now suppose there are 2 points $X,Y$ on $F$ such that $X,O,Y$ are collinear, where $O$ is the center of the circle and $OX>OY$. WLOG let $AB$ be the tangent to $F$ at $Y$, then note that $X$ is outside the rectangle $ABCD$, a contradiction, since $F$ must be in the rectangle and tangent to it. Thus for any $X,Y$ in $F$ with $X,O,Y$ collinear, we have $OX=OY$, so $F$ is symmetrical about $O$.