530
ON THE SCALENE TRANSFORMATION OF A PLANE CURVE.
[617
Reverting to the case where the locus of A is the circle
this gives
r'- — 2yr' cos 6 + 7 2 — /¿- = 0,
t = 7 cos 6 4- V(/f 2 — T sin 2 6),
1 _ 7 cos 6 — \/(/i 2 — 7 2 sin 2 6)
r 7 2 — /t 2
so that for the transformed curve we have
r 2 + r 1
in
~)yeoae + r(l +
! * 2 — 7 2 sin 2 6) + ni 2 — n 2 — 0.
Putting for shortness
ively, this is
l 2 — m 2
7 2 -IP
= X, and for r, r cos 6, r sin 6, writing ^(a? + if), x, y respect-
x 2 + y- + (1 — X) >yx + (1 + X) h 2 (x* + y 2 ) — 7 2 y 2 \ + m? — n 2 = 0,
or, what is the same thing,
[sc 2 + y 2 + (1 — X) yx + m? — n 2 ) 2 = (1 + X) 2 [IP {a? + y 1 ) — 7 2 y 2 \,
a bicircular quartic. In the case X = —1, it reduces itself to the circle
x 2 + y 2 + 2yx + m 1 — n 2 = 0
twice, which is the case considered above; and in the case X = 1, or l 2 + h? = m 2 + 7-’,
the equation is
(x 2 + y 2 + m 2 — n 2 ) 2 = 4 [h? (x 2 + y 2 ) — 7 2 y 2 },
so that the curve is symmetrical in regard to each axis. In the case 7 = 0, the locus
is a pair of concentric circles, centre B.
The equation
[x 2 + y 2 + (1 — X) yx -P m 2 — ?i 2 } 2 = (1 4- X) 2 {h 2 (x 2 + y 2 ) - 7 2 y 2 },
which contains the four constants X, 7, h and w? — n 2 , may be written in the form
(a? + y 2 4- Ax + B) 2 = ax 2 4- ey 2 ,
(where the constants A, B, a, e are also arbitrary). This is, in fact, the equation of
the general symmetrical bicircular quartic, referred to a properly-selected point on the
axis as origin, viz. the origin is the centre of any one of the three involutions formed
by the vertices (or points on the axis); say it is any one of the three involution-
centres of the curve.
To show this, assume
(x — a)(x — /3) (x — 7) (x — 8) = x? —px 3 + qxr — rx + s: