48 LECTURES AND ESSAYS [1869- 



the equation of a rectangular hyperbola through P and 



P , whose centre is the middle point of PP , and which is 

 referred to conjugate diameters inclined at angle a. The 

 orthogonal system in this case consists of the lemniscates 

 rr = c. One of the hyperbolas consists of the straight line 

 PP , and the line equidistant from P and P . Dividing 

 the plate along the latter line, we have the case of one 

 source in a plate bounded in one direction by an infinite 

 straight line, but otherwise unlimited or bounded by a 

 lemniscate of infinite conductivity, having P and its image 

 due to the boundary line for poles. 



(c) To find the image of 

 any point in a circular bound 

 ary, i.e. to find the source 

 which in combination with a 

 source at the centre of the 

 circle, and an equal sink at 

 any other point, will make 

 the circle a stream line. 

 Let A be the centre of the circle, and P the given sink. 

 In AP take P , so that AP.AP = AQ 2 . Then PAQ and 

 QAP are similar triangles, and QPA = AQP . 



Therefore QAP + QP A + QPA = 2*, or (6) is satisfied 

 for any point in the circle by assuming at P a sink = P. 



(d) Hence if there be within a circle m sources and 

 n sinks, we must assume the same number of sources 

 and sinks without the circle, and n - m sources at the 

 centre. 



(e) The straight line equidistant from two equal sources 

 of the same sign is clearly a stream line for these points. 

 Hence the image of any point in a straight line is an equal 

 point, which is its optical image. 



I have constructed the equation 



2(9 = a 



on the assumption that all the sources are equal, because 

 the degree of the stream line is equal to the number of 

 equal sources (positive and negative) to which the system 



