Quadratic by Line and Circle

Summary: Construction of linear and quadratic equations as intersections of low-order curves. A linear equation $z=ab(d-c)/(bc+ad)$ is realized by two crossing lines (§488, figure 97); the general quadratic $A{z}^2+Bz+C=0$ is realized by the intersection of a straight line and a circle (§§489–491, figure 98), and the simplest case $b=0$ reduces it to a circle on the axis with center at $z=-B/(2A)$ — geometric form of the quadratic formula.

Sources: chapter20 (§§488–491); figures 97, 98 in figures94-98.

Last updated: 2026-05-12.

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Linear case: two lines (§488, figure 97)

Take the two lines $EM$ and $FM$ meeting at $M$. Let the axis be $EF$ with origin $A$; erect the perpendicular $ABC$ at $A$, where $B$ lies on the first line and $C$ on the second. Write

$AE=a,AF=b,AB=c,AC=d,AP=z,PM=y.$

Similar triangles give the two line equations:

$\frac{a}{c}=\frac{a+z}{y}⟹ay=c(a+z),$ $\frac{b}{d}=\frac{b-z}{y}⟹by=d(b-z).$

Eliminating $y$ by $b\cdot [\text{first}]-a\cdot [\text{second}]$ yields

$bc(a+z)=ad(b-z)⟹z=\frac{ab(d-c)}{bc+ad}.$

Any linear equation $\alpha z+\beta =0$ can be brought into this form by suitable choice of $a,b,c,d$, so every linear equation admits a two-line construction.

Quadratic by line and circle (§489, figure 98)

Replace the second line by a circle. Keep $AE=a,AB=b$ as before (so the line $EM$ is $ay=b(a+z)$, i.e., $y=(ab+bz)/a$); drop the perpendicular $CD$ from the circle’s center $C$ to the axis and let $AD=f,CD=g$, radius $CM=c$. The circle’s equation is

$(z-f{)}^2+(y-g{)}^2=c^2.$

Substituting $y-g=(b-g)+bz/a$ and clearing denominators,

$(a^2+b^2)z^2+2(ab(b-g)-a^{2}f)z+a^2(b-g{)}^2+a^2{f}^2-a^2{c}^2=0.$

The two roots are the abscissas of the two intersection points $M$ and $m$ of the line and circle, perpendiculars to the axis from which give $AP$ and $Ap$.

Matching a general quadratic (§490)

To match the general $A{z}^2+Bz+C=0$, multiply through by $(a^2+b^2)/A$ and equate the coefficients of the resulting equation with the line-circle eliminant:

• coefficient of $z$: $2Aab(b-g)-2A{a}^{2}f=B(a^2+b^2)$, giving $af=b(b-g)-\frac{B(a^2+b^2)}{2Aa}.$
• constant term: $a^2(b-g{)}^2+a^2{f}^2-a^2{c}^2=\frac{C(a^2+b^2)}{A}.$

Solving for $b-g$:

$b-g=\frac{Bb}{2Aa}\pm \sqrt{\frac{a^2{c}^2}{a^2+b^2}+\frac{C}{A}-\frac{B^2}{4A^2}}.$

Three free parameters $a,b,c$ remain; they must be chosen so the radicand is positive (otherwise $b-g=AB-CD$ is complex and the construction fails).

Simplest form: $b=0$ (§491)

Setting $b=0$ (the line $EM$ coincides with the axis itself) collapses the construction. Then $f=-B/(2A)$ and $g=\sqrt{c^2-(B^2-4AC)/(4A^2)}$. The radicand is positive iff $B^2>4AC$, which is exactly the condition for $A{z}^2+Bz+C=0$ to have real roots.

The most natural choice is to make $g=0$ as well, putting the center $C$ on the axis at $D$:

$AD=-\frac{B}{2A},\text{radius}=\frac{\sqrt{B^2-4AC}}{2A}.$

The two intersections of this circle with the axis $AP$ are the two roots — this is exactly the quadratic-formula picture.

Removing the square root from the construction

To avoid having to construct $\sqrt{B^2-4AC}$ as a separate length, Euler introduces a free parameter $k$ via $g=c-k/(2A)$ and solves for $c,g$:

$c=\frac{k^2+B^2-4AC}{4kA},g=\frac{B^2-4AC-k^2}{4kA}.$

The most convenient choice is $k=-B$, giving

$AD=-\frac{B}{2A},CD=\frac{C}{B},\text{radius}=-\frac{B}{2A}+\frac{C}{B}=AD+CD.$

Center $D^{\prime }$ above $D$ at height $C/B$, radius equal to the sum $AD+CD$ — no square root needed.

Status of the quadratic construction

By the [[intersection-product-degree-bound|$mn$ bound]] a line and a circle meet in at most $1\times 2=2$ points, exactly matching the two roots of the quadratic. The alternative quadratic-by-two-circles (§492) construction also has joint intersection bound $4$ — but since two circles meet in at most $2$ points generically, it also fits the quadratic; Euler prefers the line+circle form because it has fewer free parameters and the parameters are more directly interpretable.

Figures

Figures 94–98

Related pages

• construction-of-equations
• chapter-20-on-the-construction-of-equations
• quadratic-by-two-circles
• straight-line-equation
• line-curve-intersection-bound
• intersection-product-degree-bound