Can shapes determined by number of points?

From an amazing theorem in plane curves geometry we know that vertices of triangles similar to arbitrary triangle $T$ is dense on every closed jordan curve in a plane, so if $J$ be such curve and $A,B,C$ be 3 noncollinear points on plane then at least one curve similar to $J$ contains $A,B,C$. If $J$ be a circle then exactly one circle passes through $A,B,C$. From here we reach the problem in its simplest form:

QUESTION: is circle the only shape on a Euclidean plane (not only from closed curves mentioned above) which just one similar to it defined by each set of 3 non-collinear points we consider from $\mathbb{R}^2$ ? (Means not two or more similar of the shape fits $A,B,C$, just one of it).

Note: here the mentioned "shape" can be any subset of $\mathbb{R}^2$. The "trivial triangle" trivially could not be an answer because we can find many similar of it passing through the 3 points creates it as vertices. It seems for an specified set of 3 points with predefined angles of the triangle it creates, also we may find just circle which is unique.

What about generalization to $n$ points in $\mathbb{R}^m$ which exactly define $k$ similar shapes?

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    $\begingroup$ @Gerry Myerson for equilateral triangle we have 3 two_thirds of a circle passing through its vertices, so it could not be an answer, just whole of a circle satisfies the condition not any part of it. $\endgroup$ – MasM Jul 24 '18 at 7:44
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    $\begingroup$ Please, "collinear", not "colinear". $\endgroup$ – Gerry Myerson Jul 24 '18 at 13:01
  • $\begingroup$ @GerryMyerson : odd, in French it only has one "l"; I wonder why that is $\endgroup$ – Max Jul 24 '18 at 19:41
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    $\begingroup$ @Max, the English-speaking peoples are more affluent than the French, and can better afford the luxury of extra letters. $\endgroup$ – Gerry Myerson Jul 24 '18 at 21:33

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