Parametric equation appears to generate circle and violate $x(t)^2+y(t)^2={R+}$

Question

The context of the equations is from plotting the real against the imaginary impedance of an RC circuit however the problem is entirely mathematical.

Given the two equations:

$x(w) = \frac{R}{C^2 R^2 w^2 + 1}$

$y(w) = \frac{C R^2 w}{C^2 R^2 w^2 + 1}$

Where R, C and w are positive reals.

Plotting these equations appears to yeild a arc which follows a circle. Arc shown in green, see https://www.desmos.com/calculator/m4yffxmpn6 for interactive version.

Purple - A circle centered at x=R and r=R

Red/Blue are functions x and y

Following the suggestion of this question I tried to prove this by squaring each component and summing them and seeing if the result is constant.

$x(w)^2 + y(w)^2$ $=\frac{R^2}{((RCw)^2+1)^2} + \frac{R^4 C^2 w^2}{((RCw)^2 + 1)^2}$ $=\frac{R^2}{((RCw)^2 + 1}$

This is still a function of $w$ and so not constant for all $w$ and the curve is not a circluar arc.

Can this be right? It looks so much like a circular arc that I am second guessing myself at this proof being right.

Answer 1

Let's assume your parametric equation describes (at least part of) a circle, with unknown center $(a,b)$ and radius $r$. Hence, for all $w$,

$$(x-a)^2+(y-b)^2-r^2=0$$

The first term is

$$\left(\frac{R}{R^2C^2w^2+1}-a\right)^2=\frac{(R-aR^2C^2w^2-a)^2}{(R^2C^2w^2+1)^2}=\frac{a^2R^4C^4w^4+(R-a)^2-2a(R-a)R^2C^2w^2}{(R^2C^2w^2+1)^2}$$

Then

$$\left(\frac{R^2Cw}{R^2C^2w^2+1}-b\right)^2=\frac{(R^2Cw-bR^2C^2w^2-b)^2}{(R^2C^2w^2+1)^2}\\=\frac{b^2R^4C^4w^4+R^4C^2w^2+b^2-2bR^4C^3w^3-2bR^2Cw+2b^2R^2C^2w^2}{(R^2C^2w^2+1)^2}$$

And the last is

$$\frac{-r^2(R^2C^2w^2+1)^2}{(R^2C^2w^2+1)^2}=\frac{-r^2R^4C^4w^4-r^2-2r^2R^2C^2w^2}{(R^2C^2w^2+1)^2}$$

Now, add the terms. Since we managed to get the same denominator, factor it out. When doing the addition, keep $w^k$ terms together.

The equation at the beginning becomes :

$$\frac{1}{(R^2C^2w^2+1)^2}\left[(a^2+b^2-r^2)R^4C^4w^4-2bR^4C^3w^3+(R^4C^2+2b^2R^2C^2-2a(R-a)R^2C^2-2r^2R^2C^2)w^2-2bR^2Cw+(R-a)^2+b^2-r^2\right]=0$$

This has to be true for all $w$, hence the right factor is a polynomial in $w$ with null coefficients.

Therefore:

• $a^2+b^2=r^2$
• $b=0$
• $R^2C^2(R^2+2b^2-2a(R-a)-2r^2)=0$ hence $R^2+2b^2-2a(R-a)-2r^2=0$
• $(R-a)^2+b^2-r^2=0$, hence $R^2+a^2+b^2-r^2-2aR=0$

From the first and last we get $R(R-2a)=0$ hence $a=R/2$. Put this and $b=0$ in the first and we get $r=R/2$. We can check with the third:

$$R^2+2b^2-2a(R-a)-2r^2=R^2-\frac{1}{2}R^2-\frac{1}{2}R^2=0$$

So the parametric curve is indeed an arc of the circle of center $(R/2,0)$ and radius $R/2$.

Is it the full circle? No, the circle passes through the origin, but $x$ can't be zero.

Are all other points on the circle reached? Let's see. $y$ is an odd function of $w$, so the curve is symmetric wrt the $x$ axis. For $w\to\infty$, $x\to0$, and for $w=0$, $x=R$. Therefore, all abscissas of points on the circle are reached except $0$. That is, the parametric curve, defined for $w\in\Bbb R$, is exactly the aforementioned circle minus one point at the origin. For $w>0$ you keep only points of the circle such that $y>0$ (upper half circle).