MR. R. LACHLAN ON SYSTEMS OF CIRCLES AND SPHERES. 
609 
This can only have equal roots if the coefficients h, g, f are real, and then we 
must have 
h=f=g=0, c=d=e. 
In this case one of the two spheres (x, y ) must be imaginary, and so the equation 
reduces to 
ax 2 + by~ + c(z 2 -f w 2 ■+ v 2 ) = 0, 
which clearly represents two spheres. 
Similarly, if the quintic 11 = 0 has four equal roots, and x be imaginary, the equation 
will reduce to 
ax 2 -\-by 2 -\- bz 2 -\-bw 2 -\-bv 2 =0, 
which represents the imaginary sphere x— 0. 
Let us now suppose the spheres (x, y) to be real, then let us take 2 any sphere 
cutting them orthogonally, and let ( 10 , v ) be the two points of intersection of (x, y, z ) : 
so that the equation of the absolute is 
x 2 -\-y 2 -\-z 2 — 4 101 ;, 
and the discriminating quintic H becomes 
(a + £)(/>+/') 
Jc+c, h, g 
h, d, f— 2k 
g, f—2h, e 
= 0; 
which can only have equal roots provided that 
e =g- 0, f= — 2c; 
and then the equation takes the form 
ax 2 -}- by 2 +C 2 3 -f div 2 + 2hzw —4ctcr=0. 
cl 
By taking, instead of 2=0, the sphere 2—10=0, this equation may clearly be 
reduced to the form 
ax 2 -}- by 2 -}- 2 / 1210 = 0. 
The surface represented by this equation has the point 10 = 0 for a binocle: if a or b 
is zero, the node is a unode. 
Again, if the equation H(</>+/ri/;) = 0 has four equal roots and the sphere corresponding 
to the remaining root is real : let us take this sphere as x=0, and let us take for our 
system of reference any two spheres (y, 2 ) and the two points ( 10 , v) in which the 
spheres (x, y, 2 ) intersect, so that the system is semi-orthogonal: then the equation of 
the surface must be of the form 
ax 2 -f- by 2 -|- C 2 2 + div 2 + ev 2 -f 2 fzw +2 givy + 2 / 11/2 + 2lyv + 2mzv + 2 nwv = 0 ; 
mdccclxxxvi. 4 1 
