SOLUTIONS OP PBOBLEMS. 75 



and q, and the radical axis in R. If PH be drawn perpendicular to the radical 

 axis, then 



PQ.Pq = 2AB.HP. 



COR. If the line be drawn from P to touch the circle in T, instead of 

 cutting it in Q and q, then the square of the tangent PT is equal to the 

 rectangle 2AB . HP. 



Similarly, if ph be drawn from p perpendicular to the radical axis 



pF = 2AB.hp. 



Hence, if a line be drawn touching one circle in T, and cutting the other 

 in P and p, then 



(PT) 2 : (pT? :: HP : hp. 



(2) If two straight lines touching one circle and cutting another be made 

 to approach each other indefinitely, the small arcs intercepted by their inter- 

 sections with the second circle will be ultimately proportional to their distances 

 from the point of contact. 



This result may easily be deduced from the properties of the similar 

 triangles FTP and p'pT. 



COR. If particles P, p be constrained to move in the circle A, while 

 the line Pp joining them continually touches the circle B, then the velocity 

 of P at any instant is to that of p as PT to pT; and conversely, if the 

 velocity of P at any instant be to that of P as PT to pT, then the line 

 Pp will continue to be a tangent to the circle B. 



Now let the plane of the circles be vertical and the radical axis horizontal, 

 and let gravity act on the particles P, p. The particles were projected from 

 the same point with the same velocity. Let this velocity be that due to the 

 depth of the point of projection below the radical axis. Then the square of 

 the velocity at any other point will be proportional to the perpendicular from 

 that point on the radical axis ; or, by the corollary to (l), if P and p be at 

 any time at the extremities of the line PTp, the square of the velocity of P 

 will be to the square of the velocity of p as PH to ph, that is, as (PT) 3 to 

 (pT)\ Hence, the velocities of P and p are in the proportion of PT to pT, 

 and therefore, by the corollary to (2), the line joining them will continue a 

 tangent to the circle B during each instant, and will therefore remain a tangent 

 during the motion. 



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