537] 
that is, 
SOLUTIONS OF A SMITH’S PRIZE PAPER FOR 1871. 
4.9 7 
r — a sin (6 + /3), 
which is the equation of a circle (radius = \a) passing through the fixed point. In 
fact, the point moving in such a circle with a constant velocity, moves about the 
centre with a constant angular velocity, and about any fixed point in the circumference 
with an angular velocity which is one-half of that about the centre, and is therefore 
also constant. 
Treating ¡3 as a variable parameter, to obtain the envelope we have 
0 = a cos (6 + /3), 
that is, 6 + (3 = ~ and therefore r = a, which is the equation of a circle (radius = a) 
A 
having the fixed point for its centre. This is consequently the singular solution. 
2. Determine the system of curves which satisfy the differential equation 
dx {V(l + x-) + ny) + dy {V(l + y 2 ) + nx] = 0 ; 
and show that the curve which passes through the point x = 0, y — n contains as part 
of itself the conic 
x 2 4- y 2 + 2xy V(1 + u 2 ) — n 2 = 0. 
The equation is integrable per se, viz. we have 
x \/(l + x 2 ) + log [x + V(1 + O} + y V(1 + y 2 ) + log [y + V(1 + y 2 )} + 2nxy = C, 
or, determining the constant so that for x = 0, y may be = n, 
G = n V(1 + n 2 ) + log [n + V(1 + n 2 )}, 
and the equation may be written 
sb V(1 + cc 2 ) + y V(1 + y 2 ) + 2nxy - n V(1 + n 2 ) + log fe + ~+J) 1 + = °’ 
which is evidently a transcendental curve; it may however be shown that, if 
x 2 + y 2 + 2xy V(1 + n 2 ) — n 2 = 0, 
then we have 
and 
x \J{ 1 + x 2 ) + y 1 + y 2 ) + 2nxy — n V(1 + u 2 ) = 0, 
[x + V(1 + a ‘ 2 )} [y + V(1 + y 2 )} = n + V(1 + n2 )> 
so that the equation of the curve is thus satisfied; wherefore the transcendental curve 
contains as part of itself the conic x 2 + y 2 + 2xy f(l + n 2 ) — n 2 = 0. 
[As a simple illustration as to how this may happen, take the transcendental 
curve y — sin xy = 0, which it is clear contains as part of itself the line y = 0.] 
C. VIII. 
63