<s>[INST] polynomial approximation of a circle First of all, I am not a mathematician and my mathematics are fairly rusty. I would appreciate some help
I believe there is not polynomial equation of a circle (is that right?) but take a look at this picture

In it we can see two equations. One a circle $30x^2+30y^2=300000$ and the other one a parabola   $y=120+-0.003x^2$
Obviously they are different but the parabola in the vicinity of (0,100) seems fairly parallel to the circle
This coefficients are randomly chosen but my question is if I have a circle of radius r , how can I find the coefficients of a parabola that is a bit parallel in:

*

*a point in the axis

*any point of the circle?

 [/INST] Expanding upon my comment about osculating circles ...
Suppose we have the second-degree polynomial function
$$y=ax^2+bx+c \tag1$$
that we want to be "parallel" to an origin-centered circle of radius $r$ near a point $R$ on that circle; let's say that the distance between the curves at $R$ should be $s$, and write $S$ for the corresponding point on the curve. We want to find $a$, $b$, $c$ such that the origin-centered, $(r+s)$-radius circle is the polynomial's osculating circle at $S$.
The Wikipedia entry tells us that the center of the osculating circle is given by
$$\left(\;x-y'\frac{1+y'^2}{y''}, \; y+\frac{1+y'^2}{y''}\;\right) \tag2$$ where $y'=2ax+b$ and $y''= 2a$ are the first and second derivatives of $y$. We want the center to be the origin, which gives us two conditions:
$$\begin{align}
2ax-(2ax+b)(1+(2ax+b)) &= 0 \tag3\\
1 + b^2 + 2 a c + 6 a b x + 6 a^2 x^2 &= 0 \tag4
\end{align}$$
Together with $(1)$, we can solve for $a$, $b$, $c$ to get
$$
a=-\frac{x^2 + y^2}{2 y^3} \qquad 
b= \frac{x^3}{y^3} \qquad
c= -\frac{(x^2 + y^2) (x^2 - 2 y^2)}{2 y^3} \tag{5}$$
Note that point $Q$ has coordinates $((r+s)\cos\theta,(r+s)\sin\theta)$ for some $\theta$. Substituting those into $(5)$ gives
$$a = -\frac{\csc^3\theta}{2 (r + s)} \qquad b=\cot^3\theta \qquad c= \frac12 (r + s) (2 - 3 \cos^2\theta) \csc^3\theta \tag{6}$$
so that the general form of the quadratic curve becomes
$$y = -\frac{\csc^3\theta}{2(r+s)} \left( x^2 -2x(r+s)\cos^3\theta - (r + s)^2 (2 - 3 \cos^2\theta) \right) \tag7$$
Here are some examples:
  
and a nifty animation:

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