485] 
PROBLEMS AND SOLUTIONS. 
559 
7. Considering the points F, G, H on a line and the point A as given, it has 
been seen that there are two Cartesians through A with the foci F, G, H; and the 
theorem asserts that in the circular cubics through F, G, H with the focus A, the 
foci concyclic with A lie on one or other of the two Cartesians: there are consequently 
through F, G, H with the focus A two systems of circular cubics corresponding to 
the two Cartesians respectively, each system depending upon two arbitrary parameters. 
But if we attend only to one of the two Cartesians and to the corresponding system 
of cubics, then the Cartesian passes through the four foci of each cubic, and if 
(instead of taking as given the points F, G, H and the focus A) we take as given 
the four concyclic foci А, В, C, D of a cubic, the theorem asserts that we have 
through A, B, G, D a Cartesian depending on two arbitrary parameters (or having for 
its axis an arbitrary line), and such that the foci of the Cartesian are the points of 
intersection F, G, H of its axis with the cubic. And I proceed to the proof of the 
theorem in this form. 
8. The equation of a circular cubic having four foci on the circle x 2 + y 2 — 1 = 0 is 
(x 2 + у 2 + 1) (Px + Qy) + lx 2 + 2mxy + ny- = 0; 
and this being so, the four foci are the intersections of the circle with the conic 
(Qx — Py) 2 + 2 (— nP + mQ) x + 2 (mP — IQ) у + m 2 — nl = 0. 
9. The general equation of a Cartesian is 
(x 2 + y 2 + 2 Ax + 2 By + C) 2 + 2 Dx + 2 Ey + F= 0, 
and by assuming for А, В, G, В, E, F, the following values which contain the two 
arbitrary parameters a and в, viz. by writing 
2 A = 6Q, 2B=-6P, G = a-1, В = - пв 2 Р + (тв 2 - ав) Q, 
Е = (тв 2 + ав) Р — 16 2 Q, F= — а 2 + в 2 (т 2 — nl), 
we have the equation of a system (the selected one out of two systems) of Cartesians 
through the four foci; in fact, substituting the foregoing values, the equation of the 
Cartesian is 
{x 2 + y 2 +6 (Qx - Ру) + a-l} 2 - 2ав (Qx - Py) 
+ 20 2 (— nP + mQ) x + 26 2 (mP — IQ) у — a 2 + в 2 (m 2 — nl) = 0, 
and writing herein x 2 + y 2 — 1=0, the equation reduces itself to 
в 2 {(Qx - Py) 2 + 2 (- nP + mQ) x+ 2 (mP -lQ)y + m 2 - nl\ = 0, 
verifying that the Cartesian passes through the four foci. 
The coordinates of the centre of the Cartesian are x = — A, y = —B, and the 
equation of its axis is E (x + A) - В (у + В) = 0 ; we have therefore to show that the 
points of intersection of this line with the cubic are the foci of the Cartesian.