399 



and append a single simple example in which the product admits of 

 being easily formed. 



Let the triangle C^O/), touch a circle BC in the points B and C, 

 and let the side O 2 O 3 be perpendicular to the line C^O drawn through 

 the centre of the circle, O, but be wholly exterior to the circle. Take 

 O as the origin of coordinates, and suppose that the equations to the 



circle are 



OP=x . OI+y . OJ, # 2 +y 2 =c 2 . 



Now change the origin to O', where OO'=|O 3 O 2 = b . OJ. Then 

 the equations to the circle become 



0'P=#'.OI+/.OJ, x n + (y'-b)*=c\ 

 And for the radical loci, when y' is clinant, we have (3) 



that is, the line O.O, and 



O'S=.r'. OI-M'. OJ, *' 2 -* f2 =c a , 

 that is, the rectangular hyperbola S\ S' 2 . The roots will then be 



2 R + i . 2 S\ _ 

 OJ 



and 



OJ 



OJ 



2 R+i.0 2 S' 2 _ 2 R+RM' 2 _ 2 M^ 

 OJ OJ = OJ ' 



and the factors corresponding to them in the final product in (3) 



will be 



2 3 -0 2 M\ 2 3 -0 2 M^_M^Q 3 M' 2 O, 

 OJ OJ OJ OJ ' 



