92 P. Dutt — Properties of the circle and analogous matters. [No. 2, 



Let us write the property in the form r 1 = wi?v, where r L = AP, 

 r%—BP. Describe two circles with the fixed points as centres, and 

 radii equal to a, b so that a = mb, then the equation of the circle can be 

 reduced to the form r l — a = m (r 2 — 6), which means geometrically that 

 the distance of any point on the circle r^ — mi\ from the circumferences 

 of the circles described with the fixed points as centres are in the fixed 

 ratio m. (1). 



The form r { — a=-m (r % — b) shows that the circle passes through 

 the intersections of r L = a, and r z ~ 6, and it is evident, therefore, that 

 the three circles r L = a, r%=b, and r^—mr^ will co-intersect if a — mb. 



The proposition may be enunciated geometrically as follows : — 



Let circles be described with the fixed points as centres, so that 

 their radii are in the ratio m. The circle which represents the locus 

 r 1= =w?*2 passes through the intersections of these circles. (2). 



If PT be the tangent at P we Fig. 2. 



a\ 

 cos APT ds 



have 



cos BPT d 



ds 

 dr v 



~ dr 2 



— m. 



This property can also be deduced without the use of the differential 

 calculus. GP is the normal at P. 

 sin CPA cos APT 

 •* WehaVe sin-CP^ = c-o75Pr- 

 sin CPA CA 

 sin CPA sin PCM AP 

 sin OPP' sin CPB ~ OB ' 

 sin POB BP 



CA 



But from similar triangles we can prove that 77^= m 8 (vide the 



Co 

 figure of prop. 3, Hall and Stevens, page 361). 



ai AP 



PP = m 



. cosAPT 



' ' c^PPT =W ' < 3) ' 



Let JSf, N' be the feet of the perpendiculars from T on AP and BP. 



Then PN=PT cos APT, PN' = PT cos BPT. 



.-. PN=m.PN'. 



Also AP-m.BP. 



.'. AN=m. BN', 



