537] 
SOLUTIONS OF A SMITH’S PRIZE PAPER FOR 1871. 
501 
3. Write <x = b— c, ¡3 = c —a, y = a —b; then considering the three circles and the 
three conics 
xr 
(p-ay + tf — fr, s + yfy c =l 
(x - by + y- = ~ yOL, — + w -j— = 1, 
' s ' ca K + ca 
(x-c) 2 + y 2 = - a/3, 
X 2 y 2 _ 
ab + K + ab 
where K is arbitrary; it is required to show that if a variable circle having its centre 
on one of the conics cuts at right angles the corresponding circle, the envelope is in each 
of the three cases one and the same bicircular quartic. 
x- 
Consider the circle (x — a) 2 + y 2 = — ¡3y and the conic ^ + ^^^ = 1, coor dinates 
of a point on the conic are cos 0f (be), sin 0f(K + bc), where 6 is a variable parameter; 
say for a moment these values are p and q. The equation of the variable circle is 
(x-p) 2 + (y-q) 2 = r 2 , 
and in order that this may cut at right angles the circle 
(x-a) 2 + y 2 =-/3y, 
we must have 
(p - of + q- = r 2 - (3y, 
or, substituting for r 2 its value from this equation, the equation of the variable circle is 
(x-p) 2 + (y- q) 2 = (a - pf + q 2 + ¡3y, 
that is, 
viz. this is 
x 2 + y 2 — a 2 — ¡3y — 2p (x — a) — 2qy = 0, 
(x 2 + y 2 — a 2 — ¡3y) —2 (x — a) f(bc) cos 6 — 2y f{K + be) sin 0 = 0. 
Hence taking the envelope in regard to 0, the equation is 
(x 2 + y 2 — a 2 — fiy) 2 — 4 (x — a) 2 be — 4y 2 (K + be) = 0, 
that is, 
(x 2 + y 1 — ab— ac + be) 2 — 4 (x — a) 2 be — 4y 2 (K + be) = 0, 
or, what is the same thing, 
(x 2 + y 2 ) 2 —2 (bc + ca + ab) (x 2 + y 2 ) — 4Ky 2 + 8obex 
+ b 2 c 2 + c 2 a 2 + a 2 b 2 — 2a 2 bc — 2 b 2 ca — 2 c 2 ab — 0, 
viz. this equation, being symmetrical in regard to a, b, c, is the same equation as would 
have been obtained from either of the other conics and the corresponding circle; and 
from the form of the equation it is clear that the curve is a bicircular quartic.