383] PROBLEMS AND SOLUTIONS. 57 3 
an identical equation which is easily verified. It, in fact, gives 
which is obviously true. The equation may also be written 
1, 
A» 
a 2 , 
a 4 a 
1, 
A 2 , 
A lt 
a 4 a, 
1, 
A 3 , 
A 4 , 
A 3 A. 
1, 
A 4 , 
A 3 , 
A 3 A 
= 0. 
and in this form it expresses the known theorem of the equality of the anharmonic 
ratios of A 2 , A 3 , J. 4 ) and (A 2 , A lf A 4 , A 3 ). 
But, in order to actually find the circle, we may write 
XY + LY+MX +X =0, 
A 1 + L + MA y A, + NA 2 = 0, 
A 2 +L + MAjA 2 + NA X = 0, 
A3 -P L -P At A 3 A 4 -p NA 4 = 0, 
and eliminating L, M, N, the equation of the circle is 
XY, Y, 
X , 
1 
= o, 
A 4 , 
1, 
A}A 2 , 
a 2 
a 2 , 
1, 
A\A 2 , 
A, 
A 3 , 
1, 
A 3 A 4 , 
A 4 
or, reducing, this is 
(A. 2 — A 4 ) [XY (A 3 A 4 — A^z) + ) {A 4 A 2 (A 3 + A 4 ) — A S A 4 (A 1 + A 2 )\ 
4- X (A 4 + A 2 — A 3 — ^1 4 ) + (A 3 A 4 — A 4 A^~\ = 0, 
or say 
Xy (AiA 2 — A 3 A 4 ) + Y [A 3 A 4 (A 4 + A 2 ) — A 4 A 2 (X 3 + -d 4 )| 
+ X {A 3 + A 4 — A 4 — A 2 ] -p (AjA 2 — A 3 A^) = 0: 
= 0, 
that is 
XF+1 
X 
Y