FORM OF TANGENTIAL EQUATION. 
435 
Since the equations (289) and (290) represent the same curve, their absolute terms 
must be the same. Now if the absolute term in the latter equation be reduced by 
means of the relation 
r 
+ 
— = n — , it becomes 
p , 2 / 2 ft 
(a 2 + pjp /A2 
•r 2 | 
and the absolute term in equation (289) is r n ; hence we get 
^’=(^y+(^y (29D 
That is, the sum of the squares of the radii of J and J' equals square of the distance 
between their centres. Hence J and J' cut each other orthogonally. 
137. The propositions established in articles 133-136 are those which we shall 
require for the present investigation. They are proved in the memoir already cited, 
but by another method. It is useful to recapitulate them here : — 
(1) A bicircular quartic is the envelope of a variable circle whose centre moves on a 
given conic F, called the focal conic, and which cuts a given circle J orthogonally. 
(2) The circle J is a circle of inversion of the quartic. 
(3) There are four circles of inversion and four focal conics. 
(4) The four focal conics are confocal. 
(5) The four circles of inversion are mutually orthogonal. 
(6) The centres of the circles of inversion are such that any three will form a self- 
conjugate triangle with respect to the circle which has the fourth for centre; in other 
words, the four centres form the angular points and the point of intersection of perpen- 
diculars of a plane triangle. 
138. The proposition proved in art. 131 is our fundamental one for the rectification 
of bicirculars ; it will be seen that it is a generalization 
of Mr. Roberts’s theorem already referred to. On 
account of its importance we will here give an ele- 
mentary proof of it, but under a slightly different enun- 
ciation. 
If OPP' be any line cutting a circle J in the points 
P, P', then if two circles passing through O touch J in 
the points P, P' respectively, the difference between 
their diameters is equal to the diameter of J. 
Demonstration . — Let C be the centre of J. Join CP, 
CP', and produce them to meet the line LOM drawn 
perpendicular to OP. Join PK. Now, evidently, 
PL=KM ; 
.*. P'M— PL =P'K= diameter of J. 
Hence the proposition is proved. 
