Quadratic by Two Circles

Summary: An alternative construction for the general quadratic $A{z}^2+Bz+C=0$ via the intersection of two circles (§492, figure 99). The subtraction of the two circle equations is linear, so the eliminant is again quadratic in $z$. Euler treats this as an aside — line-plus-circle (see quadratic-by-line-and-circle) is the preferred construction because it has fewer free parameters and the degree bound of two circles ($2\times 2=4$) is wasted on a quadratic.

Sources: chapter20 (§492); figure 99 in figures99-102.

Last updated: 2026-05-12.

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Setup (§492, figure 99)

Two circles in the same coordinate system. The first has center at $C$ with $AD=a,CD=b$ and radius $CM=c$; the second has center at $c$ (lowercase) with $Ad=f,dc=g$ (with $d$ below the axis in the figure) and radius $cM=h$. Writing $AP=z,PM=y$:

$\{\begin{array}{l}{z}^2-2az+a^2+y^2-2by+b^2=c^2\\ z^2-2fz+f^2+y^2+2gy+g^2=h^2\end{array}$

Elimination

Subtracting eliminates the quadratic terms in $y$ and $z$:

$2(f-a)z+a^2-f^2-2(b+g)y+b^2-g^2=c^2-h^2.$

So

$y=\frac{a^2+b^2-f^2-g^2-c^2+h^2-2(a-f)z}{2(b+g)},$

which is the radical axis of the two circles — a straight line. From $(a-x{)}^2+(b-y{)}^2=c^2$ together with this linear relation, $y$ is eliminated to give

$(4(a-f{)}^2+4(b+g{)}^2)z^2+(\text{linear in }z)+(\text{constant})=0,$

a quadratic in $z$ whose roots are the abscissas of the two intersection points $M,m$.

The full expanded eliminant (Euler’s §492):

&\bigl(4(a - f)^2 + 4(b + g)^2\bigr)z^2 \\ &\quad + \bigl(4(a - f)(c^2 - h^2) - 4(a + f)(b + g)^2 - 4(a - f)(a^2 - f^2)\bigr)z \\ &\quad + (b + g)^4 + 2(a^2 - c^2)(b + g)^2 + 2(f^2 - h^2)(b + g)^2 + (a^2 - c^2 - f^2 + h^2)^2 = 0. \end{aligned}$$ ## Why Euler relegates this to an aside Three observations: 1. The eliminant is genuinely quadratic — two circles cannot meet in more than 2 points — so the degree is *not* the joint bound $2 \times 2 = 4$. The eliminant carries no extra information beyond what the line-circle eliminant carries. 2. There are six free constants ($a, b, c, f, g, h$), three more than the line-circle setup needs. So matching to a given $Az^2 + Bz + C = 0$ leaves "an infinite number of ways" but no canonical normalization. 3. Quadratics need to be matched to *any* line-and-circle intersection by [[quadratic-by-line-and-circle|§§489–491]], which Euler has already given in canonical form (one of the two curves is the axis itself when $b = g = 0$). The two-circle construction is harder to describe and conveys no extra power. > "Since the same quadratic equation can be interpreted as the intersection of a straight line and a circle, this is the preferred interpretation, rather than two circles." (§492) ## Hard ceiling: only quadratics The construction is *fundamentally* limited to quadratics. Two circles cannot meet in more than 2 points, so no cubic or higher equation can be matched this way. To realize a quartic, Euler must move to a circle and a parabola — see [[biquadratic-by-circle-and-parabola]]. ## Figures <!-- figures: auto-embedded -->  *Figures 99–102* ## Related pages - [[construction-of-equations]] - [[chapter-20-on-the-construction-of-equations]] - [[quadratic-by-line-and-circle]] - [[biquadratic-by-circle-and-parabola]] - [[intersection-product-degree-bound]]