jV* 



224 Prof. Challis on the Solution of a Problem in the 



which so cut off is expressed by the right-hand side of this 

 equation. It is not difficult to see that this condition is fulfilled 

 by a circle of radius X, having its centre in an arbitrary position. 

 Let a. and /3 be the coordinates of the centre. Then for any 

 abscissa a?„ we have the two values ft± Vx* — («— x x ) 2 of the 

 ordinates, and the segmental area in question is the integral 

 ^ydx taken from s = s } tos=s 2 , and from ?/=/3 — ^A 2 — (a— x x ) 2 

 to y = /3 + v/X 2 — («— x^'K Conceive C to be the centre of the 

 circle, P and P' the points of intersection of the circle by the 

 ordinates, and PAP' the intercepted arc. Then if be the angle 

 which the tangent at P makes with the axis of x y we have 



- — =£Lr— =X sin (f>. So for the point P', 

 V 1 +Pi* 



— — — =Xsin (it — <£)=-Xsin <f>. 

 Vl+p 2 2 



Also, since the angle CPP' is equal to <fi, X sin <£ = «— #, for 

 both points. Hence the above equation becomes 



£ ( 5 2-*l)-i(«-^l)fe/2--yi) 



= the sector CPAP'- the triangle CPP' 



= the segment PAP'. 



From this result it may be inferred that the form of the curve 

 which satisfies the condition of a maximum is that of the circle, 

 and that the position of the centre is at disposal for fulfilling 

 required conditions. 



If it be objected that, although the investigation for the case 

 in which A 2 is less than Amc 1 led to a circle having two ordinates 

 to each abscissa, it was not necessary to take both into account 

 by entering into such considerations as the foregoing, the answer 

 is, that in that case the ordinates were equal with opposite signs, 

 and that the result would consequently have been the same if 

 the negative ordinates had been included. In fact, for this par- 

 ticular case and no other, the equation (A) is satisfied, if the 

 curve be a circle, when only one set of values of y is taken into 

 account in the integration, as may be thus shown. Integrating 

 from 5 = to s = s v and from y = to y=y 1} the result is 



and the integral is consequently the positive half of the segmental 

 area, the chord of which is equal to 2y^ ; which plainly should 

 be the case, the negative values of y having been excluded. 

 Further, it may be remarked that, if the determination of the 



