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

Author(s):
Gary Rubinstein (Stuyvesant High School)

### Constructing the circle

At this point, Descartes had parameters $$a, b,$$ and $$n$$ defining the Cartesian parabola and $$h, k,$$ and $$d$$ defining the center and radius of the circle. The final piece that bears explaining is that after Descartes created the Cartesian parabola and the center of the circle at point $$(h,k)$$ (point I on the diagram), he did something cryptic to create the circle itself.

Figure 14. Cartesian parabola and point I. Instructions: Change the values of $$p, q, r, s, t,$$ and $$u$$ to change the Cartesian parabola and the location of point I.

Rather than just say to construct a circle with a radius of length $$d=\sqrt{(a+k)^{2}+h^{2}-\frac{s-p\sqrt{u}}{n^{2}}},$$ Descartes instead created two circles, one with diameter IL where B is the origin and BL is of length $$a,$$ and another circle with center L and radius $$\sqrt{\frac{s-p\sqrt{u}}{n^{2}}}.$$ The intersection, P, of these two circles, he claimed, would be a point (or points) on the circle with center I and radius $d=\sqrt{(a+k)^{2}+h^{2}-\frac{s-p\sqrt{u}}{n^{2}}}.$

Figure 15. Two circles are used to create the main (green) circle. Instructions: Adjust the coefficients $$p, q, r, s, t,$$ and $$u$$ to see how the Cartesian parabola, three circles, and triangle change.

To see why this works, look at triangles HIL and PIL, which are both right triangles with diameter LI as their hypotenuses.

Figure 16. Construction of line segment IP. Instructions: Drag points H, I, L, and P to see that IP will have length $$\sqrt{{\rm{LH}}^2+{\rm{HI}}^2-{\rm{PL}}^2}.$$

To give IP the length $d=\sqrt{(a+k)^{2}+h^{2}-\frac{s-p\sqrt{u}}{n^{2}}},$ Descartes used the fact that, in a circle with diameter LI and points H and P on the circumference, two right triangles LPI and LHI sharing hypotenuse LI are formed. Since ${\rm{LI}}^{2}={\rm{PL}}^{2}+{\rm{IP}}^{2}={\rm{LH}}^{2}+{\rm{HI}}^{2},$ we have ${\rm{IP}}=\sqrt{{\rm{LH}}^2+{\rm{HI}}^2-{\rm{PL}}^2}.$ By making $${\rm{LH}}=a+k,$$ $${\rm{HI}}=h,$$ and $${\rm{PL}}=\sqrt{\frac{s-p\sqrt{u}}{n^{2}}},$$ IP will then have the required length $d=\sqrt{(a+k)^{2}+h^{2}-\frac{s-p\sqrt{u}}{n^{2}}}.$