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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

circle and parabola

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?
KansaiRobot
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  • For your first question, look up “Taylor polynomial” – Martin R Jan 16 '23 at 05:45
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    (You are correct that a circle is not described by a polynomial function.) A useful concept here is that of osculating ("kissing") circle, which is tangent to a curve at a given point and matches the curve's curvature there. (The matching curvature accounts for a kind of "local parallelism".) Effectively, you want the reverse: a quadratic polynomial whose osculating circle is (concentric to) a given one at a given point. – Blue Jan 16 '23 at 05:59
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    There is no polynomial function of $x$ that plots a perfect circle, but $30x^2+30y^2=300000$ actually is the polynomial equation of a circle. When we say we have an equation of a shape we don't necessarily mean that one variable is a function of the other. – David K Jan 16 '23 at 06:16

2 Answers2

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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:

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and a nifty animation:

enter image description here

Blue
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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.

Gary
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