485] 
PROBLEMS AND SOLUTIONS. 
575 
Now the tangential equation of the envelope of the line which cuts harmonically the 
last-mentioned two conics, is 
(be' + b'c - 2ff, . , . gh! + g'h ~af - af,. , .$£, y, D 2 5 
or substituting for a &c. a' &c., their values, it is found that the coefficients of this 
equation have all of them the common factor 29-, and that omitting this factor the 
equation is independent of 6, viz. the tangential equation of the envelope in question is 
(1, -k-d\ h-a\ 0, lc (2k — 1) a 2 , 0$£ y, £) 2 = 0, 
which proves the theorem. 
3. In ‘particular, if Jc = 1, that is if the points A, B lie on the conic xz—y* — 0, 
then the tangential equation of the envelope is 
that is 
(1, - a 2 , a 4 , 0, a 2 , y, = 0, 
p_ a Y + a 4 ^ 2 + 2a 2 ^=0; 
or, what is the same thing, the equation is 
(£ ~ + a 2 0 (f + ay + a 2 £) = 0, 
and thus the envelope breaks up into the two points 
£ — ay + a 2 £ = 0, £ + -f a 2 £ = 0 ; 
that is, the points (1, —a, a 2 ) and (1, a, a 2 ), which are the points A and B respectively. 
That is, in the problem in its original form, if the points A and B are the 
extremities of a diameter of a given circle, then the two constructed circles are a 
pair of orthotomic circles with the centres A and B respectively; and the theorem is 
the very obvious one, that any line through the centre of either circle cuts the two 
circles harmonically. 
[Vol. vn. pp. 52, 53.] 
2270. (Proposed by Professor Cayley.)—To reduce the equation of a bicircular 
quartic into the form SS' — k 3 L = 0, where S = 0, S' =0 are the equations of two circles, 
L = 0 the equation of a line. (See Salmon’s Higher Plane Curves, p. 128.) 
Solution by the Proposer. 
The equation of a bicircular quartic may be taken to be 
(.x 2 + y-y + (Ui + Mo) (a 2 + y~) + v 2 + v r + v 0 = 0, 
where, and in what follows, the subscript numbers denote the degrees in the coordinates 
(x, y) of the several functions to which they are attached.