160 Proceedings of Royal Society of Edinburgh. [sess. 
If we put 
c + d = a, c-d = j3 
A 4 + 4 p 2 q 2 = ft 4 , a 2 -b 2 = k 2 
the equation of the glissette is 
{ fi 4 (r 4 + a 2 /3' 2 ) - 1 §k 2 xy(px + qy)(qx -py) - 2 a/3(k^ - #)r 2 } (r 2 - a 2 )(r 2 - /3 2 ) 
+ 16{(a 2 ^ 2 + b 2 p 2 ){x 2 - y 2 ) - 2 k 2 pqxy} 2 = 0 . (6) 
h o 0.0 
ere r z = x l + y . 
If the tracing point be on either axis of the disc, the glissette 
will clearly he symmetrical about the axes of reference and also 
about the bisectors of the angles between the axes. Taking the 
tracing point on the major axis, we have q = 0, and therefore 
\i 2 = A 2 = k 2 -p 2 , so that the locus is 
{ (k 2 - p 2 ) 2 (P + a 2 /3 2 ) + 1 Qk 2 p 2 x 2 y 2 - 2 ap(k± -Y)r 2 } ( r 2 - a 2 )(r 2 - /5 2 ) 
+ \WYiY - y 2 ) 2 = o ... (7) 
If we suppose the disc reduced to a rod, we have 6 = 0, and 
therefore a = k+p, /3 = k-p, and (7) reduces to 
(d 2 x 2 + P 2 y - a 2 j3 2 )(/3 2 x 2 + a 2 y 2 - a 2 /3 2 )(r 2 - a 2 )(r 2 - J3 2 ) = 0 
as it ought to do. 
If the tracing point be at the centre of the disc, wc see from (7) 
that the glissette is 
(r 2 - c 2 ) 4 = 0 . 
In these two cases vanishes identically and the locus consists of 
the two curves A, A'. 
If the tracing point be at a focus of the disc, we have p = k and 
a 2 + j3 2 = 4a 2 , a/3 = 2b 2 , so that (6) becomes 
x 2 y 2 (r 2 - <x 2 )(r 2 - /3 2 ) + ¥(x 2 - y 2 ) 2 = 0 
i-e, 
x 2 y 2 {P — 4a 2 r 2 + 46 4 } + 6 4 (r 4 - 4 x 2 y 2 ) = 0 
or 
r 2 {r 2 (x 2 y 2 + 6 4 ) - 4 a 2 x 2 y 2 ) = 0 
Thus the locus consists of the sextic 
+ 6 4 ) = 4 a 2 x 2 y 2 
