of Edinburgh, Session 1869-70. 83 
same thing, sinks at an infinite distance, the stream lines are 
rectangular hyperbolas. For in this case, 
P P r N x 
QPN + QP'N = a = QNx, if we make P'QN = QPN. Also QN 
touches the circle through PP'Q, therefore 
QN 2 = NP' . N P 
- the equation of a rectangular hyperbola through P and P', 
whose centre is the middle point ofPP', and which is referred 
to conjugate diameters inclined at angle a. The orthogonal 
system in this case consists of the lemniscates rr f - c. One of 
the hyperbolas consists of the straight line PP', and the line equi- 
distant 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 boundary, 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 4- QPA = 2tt, or (6) is satisfied for any 
point in the circle by assuming at P' a sink == P. 
(rZ.) Hence if there be within a circle m sources and n sinks, we 
