94 
Proceedings of the Royal Society 
It is easily shown that the stream lines orthogonal to these are 
lemniscates with the same centre, passing through the four points, 
one of which becomes a circle when the parallelogram is rectan- 
gular. 
The ellipse appears to be an impossible conic for four points, 
for conics occur in pairs orthogonal to each other. The orthogonal 
of the ellipse must be a confocal hyperbola, which is impossible, 
the only hyperbola being that discussed above. Orthogonal circles, 
however, are possible, and fall under two classes, according as all 
the points are on one circle, or two on each. 
If ABCD lie on a circle, that circle is obviously a stream line. 
Let BA.DC produced meet in 0. Then OA.OB = 00. OD, and 
the circle, with centre 0 and radius ^/OA.OB is the other branch 
of the stream line. If 0 lies within the circle ABCD, the second 
circle becomes impossible. If CA.BD produced meet B and 
CB.AD in S, B and S are centres of equipotential circles, only 
one of which is real, unless the second stream circle is imaginary. 
We may take as an example the case of a rectangle, points of the 
same sign lying on the same diagonal. Let the circle through the 
four points be (2a and 2b being the sides of the rectangle) 
x 2 + y 2 - a 2 - b 2 = 0 . 
The other branch is the imaginary circle 
x 2 + y 2 + a 2 + b 2 — 0 ; 
and we know that another stream line is the hyperbola 
y 2 - x 2 - a 2 + b 2 = 0 . 
Hence the stream lines are 
(x 2 + y y - (a 2 + b 2 ) 2 + \(y 2 - x 2 - a 2 + b 2 ) = 0 , 
lemniscates as above. 
The equipotential circles degenerate into the straight lines 
x — 0 and y = 0 . 
If 0 be the point in OD produced which is equidistant from A 
and B, and OC.OD = OA 2 = OB 2 , the circle with 0 as centre 
passing through A,B is a line of flow. 
The circle having its centre P in AB produced, and passing 
through CD, is obviously orthogonal ; and since PA.PB = PC 2 
