606 
PROBLEMS AND SOLUTIONS. 
[383 
and secondly, 
that is, 
Ci7i, 
1, 
«i, 
A 
= 
7i, a i, 
A, 
7! 
1, 
a 2 , 
A 
72, «2, 
A, 
72 
C 3 73> 
1, 
«3, 
A 
78, «3, 
A, 
73 
C474» 
1, 
a 4 , 
A 
0^4? ^4> 
A, 
74 
1, 
«n 
A, Ci 7! 
= 0; 
1, 
a 2, 
A, C 2 7-_, 
1, 
a 3 , 
A, c 3 7 3 
1, 
«4, 
A, c 4 7 4 
may be 
united 
into the single 
formula 
1, 
«1, 
A 
, Cj + 7!, c 1 y 1 
= 0. 
1, 
a,, 
fi-2, 
C2 + 7.., 
'272 
1, 
a-j, 
A 
. c 3 4- 7 3 , c 3 7 ;i 
1, 
«4, 
A 
C4 + 74, C474 
Now the equation of a sphere having for the extremities of a diameter the points 
(a, /3, 7) and (a, b, c) is 
O - £0 + a)] 2 + [y -\(b + /3)] 2 + [z-$(c + 7 )] 2 = l [(a - a) 2 + (b -/3) 2 + (c - y) 2 ], 
or 
o - a) (x - a) + (y - b) (y-fi) + (z- c) (z-y) = 0, 
or 
x 2 + y- + z 2 — (a + a) x — (b + /3) y — (c + 7) 2 + aa + b/3 + c<y = 0; 
therefore, when the two points are (a, /3, 7) and (0, 0, c), the equation is 
x 2 + y 2 + z 2 — olx — (3y — (c + 7) £ + cy = 0. 
Hence, putting for shortness P = — ax — /3y—(c+^z+cy, viz., Pj = — a^x—{3 x y - {c 1 -\-y l )z+c l y 1 , 
&c., the equations of the four spheres are 
x 2 + y 1 + z- + Pi = 0, x 2 + y 2 + z 2 + P 2 = 0, x 2 + y- + z 2 + P 3 = 0, x 2 + y 2 + z n - 4- P 4 = 0, 
and the four spheres will have a common radical axis, if for proper values of the 
multipliers y, v, p we have 
y (P, -P 2 ) + v (P x - Pa) + P (Pi P 4 ) = 0, 
or what is the same thing, if for proper values of X, y, v, p we have 
XP ! + yP 2 + vP s + pP 4 — 0, 'h. J ry-\'V J rp — 0 ;