By saying circle-tangent polynomial I mean a polynomial of degree $N>2$ that lies mostly inside a unit circle except for the two tails exiting the circle and going to infinity and is tangent to the unit circle at all of its $N-1$ points of intersection with that circle, as in the graphs below:
As shown in the graph all such polynomials seem to be either odd or even functions. I can even show that a circle-tangent cubic must only be in the form $y=Ax^3+Cx$ as follows.
My attempt to prove that circle-tangent cubics must be odd
Consider $y=Ax^3+Bx^2+Cx+D,A\ne 0$. We can find the points of intersections of the cubic and the unit circle with
$$x^2+(Ax^3+Bx^2+Cx+D)^2-1=0,$$
which is a polynomial of degree 6.
However, noticing that the cubic must be tangent to the circle at all intersection points, all of its distinct roots must be repeated roots. In fact, since the two tails that lead to infinity crosses the circle while being tangent at the intersections, there must be two distinct roots, each with a multiplicity of 3. Naming them $p$ and $q$, we can rewrite the resultant sextic with
$$\begin{align}x^2+(Ax^3+Bx^2+Cx+D)^2-1\equiv k(x-p)^3(x-q)^3&&(1)\end{align}$$
Replacing $x$ with $-x$ we have
$$\begin{align}x^2+(-Ax^3+Bx^2-Cx+D)^2-1\equiv k(x+p)^3(x+q)^3&&(2)\end{align}$$
$(1)-(2)$, we have
$$4(Ax^3+Cx)(Bx^2+D)\equiv k\left((x-p)^3(x-q)^3-(x+p)^3(x+q)^3\right).$$
Plugging $x=p$, we have
$$\begin{align}(Ap^3+Cp)(Bp^2+D)\equiv -2kp^3(p+q)^3&&(3)\end{align}$$
and plugging $x=q$, we have
$$\begin{align}(Aq^3+Cq)(Bq^2+D)\equiv -2kq^3(p+q)^3&&(4)\end{align}$$
Assuming $p,q\ne 0$, and we notice that on the right hand side $\frac{(3)}{p^3}\equiv \frac{(4)}{q^3}$, so
$$ \begin{align} \frac{(Ap^3+Cp)(Bp^2+D)}{p^3}&\equiv \frac{(Aq^3+Cq)(Bq^2+D)}{q^3}\\ \left(Ap+\frac Cp\right)\left(Bp+\frac Dp\right)&\equiv \left(Aq+\frac Cq\right)\left(Bq+\frac Dq\right)\\ ABp^2+\frac{CD}{p^2}&\equiv ABq^2+\frac{CD}{q^2} \end{align} $$
In order to make the identity hold, the only conclusion is that $p=-q$ (since $p\neq q$) or $AB=CD=0$.
If $p=-q$, then $(1)$ can be rewritten as
$$x^2+(Ax^3+Bx^2+Cx+D)^2-1\equiv k(x^2-p^2)^3$$
Since the right hand side is even, so does the left hand side. This gives that the cubic must be odd.
On the other hand, if $AB=CD=0$, then obviously $B=0$. With this, the $x^5$ term of the resultant sextic disappears, and according to Vieta's theorem $3p+3q=0$, giving $p=-q$, and from above we have $D=0$ as well.
Generalizing the argument to higher powers
Since for a degree-$N (N>2)$ circle-tangent polynomial $P_N(x)=\sum_{n=0}^N A_nx^n$ there are $N-1$ intersection points, two of which have multiplicity 3 while the remaining have multiplicity 2, we should be able to rewrite $x^2+(P_N(x))^2-1$ with the identity
$$\begin{align}x^2+(P_N(x))^2-1\equiv k(x-r_1)^3\left(\prod_{n=2}^{N-2}(x-r_n)^2\right)(x-r_{N-1})^3&&(5)\end{align}$$
The argument above still works for the step replacing $x$ with $-x$:
$$\begin{align}x^2+(P_N(-x))^2-1\equiv k(x+r_1)^3\left(\prod_{n=2}^{N-2}(x+r_n)^2\right)(x+r_{N-1})^3&&(6)\end{align}$$
$(5)-(6)$ gives
$$4\left(\sum_{m=0}^{\lfloor{N/2}\rfloor}A_{2m}x^{2m}\right)\left(\sum_{n=0}^{\lfloor{(N-1)/2}\rfloor}A_{2n+1}x^{2n+1}\right)\equiv k\left((x-r_1)^3\left(\prod_{n=2}^{N-2}(x-r_n)^2\right)(x-r_{N-1})^3-(x+r_1)^3\left(\prod_{n=2}^{N-2}(x+r_n)^2\right)(x+r_{N-1})^3\right)$$
which, if plugging $x=r_1,r_2,\cdots,r_{N-1}$, loses the symmetry it has in the cubic case. And here is where I stopped.
I would like to ask, if there is any method to extend my argument that such circle-tangent polynomials must be either odd or even to higher order, or if there can be any counterexamples for higher orders.
