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:

- \(p=-2b\)
- \(q=n^{2}-2kn+b^{2}-2an\)
- \(r=-2hn^{2}+2bkn+4abn\)
- \(s=-n^{2}d^{2}+k^{2}n^{2}+h^{2}n^{2}+2akn^{2}-2ab^{2}n+a^{2}n^{2}\)
- \(t=-2abkn^{2}-2a^{2}bn^{2}\)
- \(u=a^{2}b^{2}n^{2}\)

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 \(u.\)

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

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 \[n=\sqrt{q-\frac{t}{\sqrt{u}}-\frac{p^{2}}{4}}.\]

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

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 \(k =-\frac{t}{2n\sqrt{u}}+\frac{2\sqrt{u}}{pn}.\)

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 \[h=-\frac{r}{2n^{2}}+\frac{\sqrt{u}}{n^{2}}+\frac{pt}{4n^{2}\sqrt{u}}.\]

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.