572 
PROBLEMS AND SOLUTIONS. 
[383 
for the coordinates of the one point; and similarly 
x — iy = a 1 — i(3 1 , x + iy = a 2 + i/3. 2 , 
for the coordinates of the other point. 
Hence, taking the new coordinates 
X = x + iy, Y = x — iy, 
and similarly A l = a l + ifi x , B 1 = cl 1 — ij3 x , &c.; the coordinates of the antifocal points V, 2' 
are (A 1} B 2 ) and (AB x ) respectively; but we have A 1 B 1 = o^ 2 + /3 2 = 1, A 2 B 2 = a 2 2 + B 2 = 1 ; 
so that Bi = -^~, B. 2 = and the coordinates are ^, {a 2 , ^ respectively. 
Similarly the coordinates of the antifocal points (3', 4') are respec 
tively. 
Take as the equation of the circle through the two pairs of antifocal points 
x 2 + y 2 4- 2\x + 2 yay + v = 0, 
or, what is the same thing, 
that is 
if 
We ought then to have 
XY+\(X + Y)-in(X- Y) + v=0, 
XY + LY+MX + X=0, 
L = \ + iy, M =\ — i/JL, N — V. 
^ r 1 + L-±- + MA 1 + N = 0, 
^ + +ma 2 + x=o, 
■“l -Ai 
^■+L±-+MA Z + N=H, 
il 4 Xi 4 
~+L l~+MA i + N = 0; 
^3 -^3 
and these will exist simultaneously, if 
A 
1 
AY 
a 2 
a 2 
1 
AY 
A x 
A 3 
1 
AY 
à; 
A 4 
1 
AY 
A 3 
A lt 1 
A 2 , 1 
A 3 , 1 
A 4 , 1 
= 0,