I want to derive that $x^2 +y^2 -r^2 = 0$ must be the implicit equation for the shape of a circle by usage of symmetries. My idea is inspired from how the roots of a polynomial completely determine the polynomial upto scaling, so to begin I consider the implicit equation of circle as : $$C(x,y)=0$$
Now, suppose we take a point $(x,y)$ on the circle then this means that new points generated by rotating original point by any angle $\theta$ would solve the implicit equation
$$ \begin{bmatrix} x \cr y \end{bmatrix} \xrightarrow[]{\text{rotation by } \theta }\begin{bmatrix} x \cos \theta -y \sin \theta \cr y \cos \theta + x \sin \theta\end{bmatrix}$$
Hence, $$ C(x,y) =0\implies C(x \cos \theta - y \sin \theta, y \cos \theta + x \sin \theta)=0$$
Now.. I'm not sure how to get the functional representation...
An alternate approach:
I came up with another way to do with actually works but it first involves assuming that it is some general conic:
$$C(x,y) = Ax^2 + Bxy +Cy^2 + Dy +Gx + H=0$$
Now, I considered transformations which still keeps me on the circle, for example reflection of a point $(x,y)$ about the line $y=x$ is a solution , hence $(y,x)$ is a solution to curve:
$$C(y,x) = Ay^2 + Byx + Cx^2 + Dx + Gy + H = 0$$
Similarly we can observe that if $C(x,y)$ is a solution to the equation, then $C(-x,-y)$ is also a solution (from the figure):
$$C(-x,-y) = Ax^2 +Bxy+Cy^2 -Dy -Gx + H=0$$
And also, we can see that reflecting our point about $x$ and $y$ axis keeps on circle so : $$ C(-x,y)=Ax^2 -Bxy+Cy^2+Dy-Gx+H=0$$ $$ C(x,-y)=Ax^2 -Bxy + Cy^2 - Dy +Gx+H=0$$
Now, composing any of that with a reflection along $y=x$ line ( I choose $C(x,-y)$:
$$C(-y,x) = Ay^2 - Byx + Cx^2 + Dx - Gy + H = 0$$
And now we have six equations and six unknowns, writing in matrix form:
$$ \begin{bmatrix} A&B&C&D&G&H \cr C&B&A&G&D&H \cr A&B&C&-D&-G&H \cr A&-B&C&D&-G&H \cr A&-B&C&-D&G&H \cr C&-B&A&-G&D&H\cr\end{bmatrix} \begin{bmatrix} x^2 \cr xy \cr y^2 \cr y \cr x \cr 1 \end{bmatrix}= 0$$
Solving this system we will get the conditions for second degree curve to be a circle.
My question: I am looking if it was possible to derive the equation in first way directly without having such an algebra mess.
You cannot get an explicit form for $C$ using your first argument, since any relation of the form $f(x^2+y^2)=c$ Also defines a circle for nice $f$. You are requiring that a certain level set of $C$ be rotationally invariant but this is satisfied by any radial function and by non radial ones having a circular $0$-level set