PROBLEMS AND SOLUTIONS. 
574 
[383 
which is the required equation; or, transforming to the original axes, we have x + iy — X, 
x — iy = F, &c., and therefore XY = x 2 + y' 2 ; and the equation becomes 
x 2 + y 2 4-1 , x + iy 
x — iy 
= 0, 
+ «2 + i (ßi + ß 2 ), (cq + ißß (a2 + î/3 2 ), 1 
a 3 + a 4 + i (ß 3 + ß 4 \ (a 3 + iß 3 ) (a 4 + iß 4 ), 1 
which is the equation of the circle through the two pairs of antifocal points. 
{Note. The second form of the equation of the circle may be otherwise deduced 
from the first, without expanding the determinants, by the following method: 
XY, 
Y, 
Z, 
1 
= 
ZF+1, 
Y, 
X, 
1 
A i, 
1, 
djdo, 
d 2 
-di + A 2 , 
1 , 
-di-d 2 , 
A 2 
d 2 , 
1, 
d]d 2 , 
A l 
A 1 + d 2 , 
1, 
d-l-do, 
A4 
dg, 
1, 
-d.3-d.4i 
d-4 
A 3 + d 4 , 
1, 
d 3 d 4 , 
d-4 
ZF+1, F, Z, 1 
= (d4-d 2 ) 
ZF+1, Z, F 
dj + d 2 , 1, did,, d 2 
dj + d 2 , d. 4 d 2 , 1 
0, 0, o, dj - d 2 
d 3 + d 4 , d 3 A 4 , 1 
d 3 + A 4 , 1, d 3 d 4 , d 4 
therefore 
XY+ 1, 
Ai + d 2 , 
A 3 + d 4 , 
X, 
did 2 , 
d 3 d 4 , 
Ed. [W. J. M.]} 
[Vol. II. pp. 22—24.] 
1513. (Proposed by the Rev. J. Blissard, B.A.)—Prove the following formulas 
(x — 1) (x — 2).. (x — n) 
(1) 
X (x + 1) .. (x + n — 1) 
l+(-\nJ n 1 n(n 2 -P) 1 n(n 2 — l 2 )(n 2 — 2 2 ) 
1+ ' ' > 'x P ' x+l + P. 2 2 'x+2