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Classical

Radical Axes and Orthogonal Parabolas

Two short proofs from quadratic equations

Subtracting circle equations produces the radical axes; adding normalized equations of two parabolas with perpendicular axes produces a circle through their four intersections.

Methods Quadratic equations · Linear combinations · Radical axis · Cartesian coordinates

The radical-axis theorem

Theorem

If three circles have three distinct pairwise radical axes, then those axes are concurrent or parallel.

Three nonconcentric circles with their three pairwise radical axes meeting at one point.
Figure 1. The three pairwise radical axes are concurrent.

Proof

Write the three circle equations in the form

Γi:Fi(x,y)=x2+y2+aix+biy+ci=0(i=1,2,3).\Gamma_i:\quad F_i(x,y) = x^2+y^2+a_i x+b_i y+c_i = 0 \qquad (i=1,2,3).

For each pair, subtracting the two equations eliminates the common quadratic part. Thus

L12=F1−F2,L23=F2−F3,L13=F1−F3L_{12}=F_1-F_2,\qquad L_{23}=F_2-F_3,\qquad L_{13}=F_1-F_3

are linear expressions, and the corresponding equations Lij=0L_{ij}=0 are the three radical axes.

But

L13=L12+L23.L_{13}=L_{12}+L_{23}.

If the first two axes meet, their intersection therefore lies on the third. If the first two axes are parallel, their linear parts are proportional, and the identity shows that the third has the same direction. Hence the three radical axes are concurrent or parallel.

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Four intersections of parabolas with perpendicular axes

Proposition

If two parabolas with perpendicular axes intersect at four distinct points, then those four points are concyclic.

Two parabolas with perpendicular axes meeting in four points on a common circle.
Figure 2. The parabolas x² - 2y - 1 = 0 and y² - x - 2y - 1 = 0 have perpendicular axes, and their four common points lie on x² + y² - x - 4y - 2 = 0.

Proof

Choose orthonormal coordinates parallel to the two axes. After interchanging the coordinate names if necessary and multiplying each equation by a nonzero constant, the parabolas can be written as

P1:x2+ax+by+c=0,\mathcal P_1:\quad x^2+a x+b y+c=0,P2:y2+dx+ey+f=0.\mathcal P_2:\quad y^2+d x+e y+f=0.

Every common point of the two parabolas satisfies the sum of these equations:

x2+y2+(a+d)x+(b+e)y+(c+f)=0.x^2+y^2+(a+d)x+(b+e)y+(c+f)=0.

This is the equation of a circle. Hence all four common points are concyclic.

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In the first argument, equal quadratic parts cancel; in the second, complementary quadratic parts combine to form x2+y2x^2+y^2.