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Can shapes determined by some 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 an Euclidean plane (not only from closed curves mentioned above) which just one similar to it to be 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 for a refined question we take the mentioned "shape" as one dimensional connected curve. The "trivial triangle" trivially could not be an answer because we can find infinitely many similar of it passing through the 3 points which creates it as vertices.

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
    Commented Jul 24, 2018 at 7:44
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    $\begingroup$ Please, "collinear", not "colinear". $\endgroup$ Commented Jul 24, 2018 at 13:01
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    $\begingroup$ @GerryMyerson : odd, in French it only has one "l"; I wonder why that is $\endgroup$ Commented Jul 24, 2018 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. $\endgroup$ Commented Jul 24, 2018 at 21:33
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    $\begingroup$ @GerryMyerson My preferred interpretation is that there's only one letter-l, but we also include a vertical line to remind students what it means. $\endgroup$ Commented Feb 22, 2022 at 5:57

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