200 Methods for the Solution of Special Problems [OH. vm 



A special investigation is needed 

 when the sphere passes through 0. Let 

 08 be the diameter through 0, and let 

 S' be the point inverse to S. Then, if 

 P f is the inverse of any point P on the 

 circle, 



OP. OP =08. 08', 



gp__qs f 



08 ~ OP ' 



or 



so that POS, S'OP* are similar triangles. 

 Since OPS is a right angle, it follows 

 that OS'P' is a right angle, so that the 

 locus of P' is a plane through 8' perpen- 

 dicular to 08'. Thus the inverse of a FIG. 69. 

 sphere which passes through the centre 



of inversion is a plane, and, conversely, the inverse of any plane is a sphere 

 which passes through the centre of inversion. 



228. If P, Q are adjacent points on a surface, and P', Q' are the corre- 

 sponding points on its inverse, then OPQ, 



OQ'P' are similar triangles, so that PQ, 

 P'Q' make equal angles with OPP'. By 

 making PQ coincide, we find that the 

 tangent plane at P to the surface PQ 

 and the tangent plane at P' to the sur- 

 face P'Q' make equal angles with OPP'. 

 Hence, if we invert two surfaces which 

 intersect in P, we find that the angle 



between the two inverse surfaces at P' is equal to the angle between the 

 original surfaces at P, i.e. an angle of intersection is not altered by inversion. 



Also, if a small cone through cuts off areas dS, dS' from the surface 

 PQ . . . and its inverse P'Q' . . . , it follows that 



dS _ OP 2 

 dS'~ OP' 2 ' 



Electrical Applications. 



229. Let PP', QQ' be two pairs of inverse points (fig. 70). Let a charge 

 e at Q produce potential V P at P, and let a charge e at Q' produce potential 

 Vp at P', so that 



FIG. 70. 



_i_ v 



PQ' Vp 



P'Q" 



then 



V P ~e'P'Q e'OQ" 



