60 LECTURES AND ESSAYS [1869- 



cussed 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 A B C D lie on a circle, that circle is obviously a 

 stream line. Let BA.DC produced meet in O. Then 

 OA.OB = OC.OD, and the circle, with centre O and radius 

 \/OAXXB is the other branch of the stream line. If O lies 

 within the circle ABCD, the second circle becomes 

 impossible. If CA.BD produced meet R and CB.AD in 

 S, R 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 (20 and 26 being the 

 sides of the rectangle) 



x &amp;lt;2 + y2_ a 2_tf = Q. 



The other branch is the imaginary circle 



x 2 +y 2 + a* + 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 2 ) 2 - (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 = and y- 0. 



If O be the point in CD produced which is equidistant 

 from A and B, and OC.OD = OA 2 = OB 2 , the circle with 

 O 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 = PD 2 is also a line of flow. In this case 

 both circles are necessarily real. 



It is clearly impossible that the same system should 



