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?

 A: Note that your circle is just
$$
x^2  + y^2  = 100^2 ,
$$
a circle with radius $100$ centered at the origin. For the upper half circle then
$$
y = \sqrt {100^2  - x^2 }  = 100\sqrt {1 - \left( {\frac{x}{{100}}} \right)^2 }  \approx 100\left( {1 - \frac{1}{2}\left( {\frac{x}{{100}}} \right)^2 } \right) = 100 - \frac{{x^2 }}{{200}},
$$
using Taylor approximation about $x=0$. In general for $x^2+y^2=r^2$, you will get
$$
y \approx r - \frac{{x^2 }}{{2r}}.
$$
This approximation is good provided $|x| \ll r$. Look up Taylor polynomials or the binomial series.
A: 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:

