Descartes’ Method for Constructing Roots of Polynomials with ‘Simple’ Curves - Derivation of Descartes' Method: Parameters

Finding relationships between parameters using a system of six equations and six unknowns

From the equation of the Cartesian parabola, $y=\frac{x^{2}}{n}-\frac{bx}{n}-a+\frac{ab}{x},$ and knowing already what Descartes’ results were, I tackled the challenge of re-deriving Descartes’ parameters for his construction – starting with the coefficients of a sextic equation to come up with the center and radius of the circle and also of the two parameters that define the Cartesian parabola so that the $x$-intercepts of the intersections would equal the real roots of the sextic equation.

The Cartesian parabola has three parameters, $n$ (the coefficient of $y$ in the original sliding parabola), $b$ (the fixed $x$-intercept of the line that intercepts the sliding parabola), and $a$ (the $y$-intercept of the line, which is always the same distance above the vertex of the sliding parabola). The circle has three parameters, $h$ (the $x$-coordinate of the center of the circle), $k$ (the $y$-coordinate of the center of the circle), and the radius of the circle, $d.$ The goal is to relate these to the six coefficients of the sextic equation $p, q, r, s, t,$ and $u.$

To find these relations, substitute $y=\frac{x^{2}}{n}-\frac{bx}{n}-a+\frac{ab}{x}$ into the equation $(x-h)^{2}+(y-k)^{2}=d^{2}.$ Combine like terms and multiply through by $x^{2}n^{2}$ to get the equation:

$$x^{6}-2bx^{5}+(n^{2}-2kn+b^{2}-2an)x^{4}+(-2hn^{2}+2bkn+4abn)x^{3}$$

$$+(-n^{2}d^{2}+k^{2}n^{2}+h^{2}n^{2}+2akn^{2}-2ab^{2}n+a^{2}n^{2})x^{2}$$

$$+(-2abkn^{2}-2a^{2}bn^{2})x+a^{2}b^{2}n^{2}=0.$$

Equate coefficients with those of $$x^{6}+px^{5}+qx^{4}+rx^{3}+sx^{2}+tx+u=0$$ to get the system of equations:

1.  $p=-2b$
2.  $q=n^{2}-2kn+b^{2}-2an$
3.  $r=-2hn^{2}+2bkn+4abn$
4.  $s=-n^{2}d^{2}+k^{2}n^{2}+h^{2}n^{2}+2akn^{2}-2ab^{2}n+a^{2}n^{2}$
5.  $t=-2abkn^{2}-2a^{2}bn^{2}$
6.  $u=a^{2}b^{2}n^{2}$

From these six equations in six unknowns, the goal is to somehow solve for $a, b, d, n, h,$ and $k$ in terms of $p, q, r, s, t,$ and $u.$

Solving for b

Divide both sides of equation (1) by $-2$ to get $b=-\frac{p}{2}.$

Solving for n

Equation (2) can be rearranged to read $n^{2}=q+2kn+2an-b^{2}$ or $n={\sqrt{q+2kn+2an-b^2}}.$ From equation (6), ${\sqrt{u}}=abn.$ Substituting $abn = {\sqrt{u}}$ into equation (5) yields ${\frac{t}{\sqrt{u}}}=-2kn-2an$ and $-{\frac{t}{\sqrt{u}}}=2kn+2an,$ which are equivalent to the second and third term under the radical in the rearrangement of equation (2). Also, since $b=-\frac{p}{2}$ so that $b^2={\frac{p^2}{4}},$ we have $$n=\sqrt{q-\frac{t}{\sqrt{u}}-\frac{p^{2}}{4}}.$$

Solving for a

That $a=\frac{2abn}{2bn},$ $abn=\sqrt{u},$ and $2b=-p$ yields $a=\frac{2\sqrt{u}}{-pn},$ or $a=-\frac{2\sqrt{u}}{pn}.$

Solving for k

Solve for $k$ in equation (5), $t=-2abkn^{2}-2a^{2}bn^{2},$ to get $k=\frac{t}{-2n(abn)}-a.$ Substitute $abn=\sqrt{u}$ and $a=-\frac{2\sqrt{u}}{pn}$ into this equation to obtain $k =-\frac{t}{2n\sqrt{u}}+\frac{2\sqrt{u}}{pn}.$

Solving for h

Solve for $h$ in equation (3), $r=-2hn^{2}+2bkn+4abn,$ to get $h=-\frac{r}{2n^{2}}+\frac{2abn}{n^{2}}+\frac{2bkn}{2n^{2}}.$ Substitute $abn=\sqrt{u}$ and $2b=-p$ into this equation to obtain $h=-\frac{r}{2n^{2}}+\frac{2\sqrt{u}}{n^{2}}-\frac{pkn}{2n^{2}}.$ This last equation can be rewritten as follows:

$h=-\frac{r}{2n^{2}}+\frac{\sqrt{u}}{n^{2}}+\frac{\sqrt{u}}{n^{2}}-\frac{2pkn\sqrt{u}}{4n^{2}\sqrt{u}} = -\frac{r}{2n^{2}}+\frac{\sqrt{u}}{n^{2}}+\frac{4u-2pkn\sqrt{u}}{4n^{2}\sqrt{u}}.$

Substitute $\sqrt{u}=abn$ and $p=-2b$ back into the numerator of the third term to get:

$h=-\frac{r}{2n^{2}}+\frac{\sqrt{u}}{n^{2}}+\frac{4a^{2}b^{2}n^{2}-2(-2b)knabn}{4n^{2}\sqrt{u}}=-\frac{r}{2n^{2}}+\frac{\sqrt{u}}{n^{2}}+\frac{-2b(-2a^{2}bn^{2}-2abkn^{2})}{4n^{2}\sqrt{u}}.$

Finally, substitution of $-2b=p$ and equation (5), $-2abkn^{2}-2a^{2}bn^{2}=t,$ into the third term yields $$h=-\frac{r}{2n^{2}}+\frac{\sqrt{u}}{n^{2}}+\frac{pt}{4n^{2}\sqrt{u}}.$$

Solving for d

Solve for $d^2$ in equation (4), $s=-n^{2}d^{2}+k^{2}n^{2}+h^{2}n^{2}+2akn^{2}-2ab^{2}n+a^{2}n^{2},$ to get:

$d^{2}=-\frac{s}{n^2}+k^{2}+h^{2}+2ak-\frac{2ab^{2}}{n}+a^{2}=(a+k)^{2}+h^{2}-\frac{s+2a{b^2}n}{n^{2}}.$

Substitute $abn=\sqrt{u}$ and $2b=-p$ into the last term to obtain $d^{2}=(a+k)^{2}+h^{2}-\frac{s-p\sqrt{u}}{n^{2}},$ or

$$d=\sqrt{(a+k)^{2}+h^{2}-\frac{s-p\sqrt{u}}{n^{2}}}.$$

At this point, Descartes had all six parameters $a, b, n, h, k,$ and $d$ expressed in terms of the coefficients of the sextic equation $p, q, r, s, t,$ and $u.$ Here, $a, b,$ and $n$ define the Cartesian parabola and $(h,k)$ and $d$ define the center and radius of the circle.

The final piece that bears explaining is how, after Descartes created the Cartesian parabola and the center of the circle at point $(h,k),$ he created the circle itself.