14 



AN ACCOUNT OF TEE EARLY MATHEMATICAL AND 



tinued for several years to propose variations of the original 

 problem, in the hope that some fortunate result would be 

 developed which would enable him to silence his opponents. 

 Mr. Dalton, however, does not appear to have taken any 

 part in the dispute : — he was ever averse to controversy under 

 any circumstances, and although a confirmation of his results 

 was offered by the editor in the Diary for 1799, he did not 

 choose to enter the lists in favour of the erroneous principle. 

 His solution of question 7 in this series may be selected as 

 affording a neat specimen of his method of treating a 

 geometrical locus. 



Question 606. By Mr, John Fletcher, " If through 

 any point P in the periphery of a circle, that is wholly 

 included in another, an indefinite number of right lines be 

 drawn to cut the periphery again in R, and terminate in the 

 circumference of the greater circle in E and F, and from E 

 towards F there be always taken EL = RF; required the 

 locus of the point L." 



Answered hy Mr, John Dalton, 

 C and O let fall CD and OB 

 perpendicular to any right line 

 FPE drawn as per question. 

 Then since DR = DL, and 

 BR=BP, therefore PL = 2BD; 

 through P draw a line parallel 

 to OC, on which set off PG = 

 20C; then will G be a point in 

 the curve the farthest possible 

 from P. Join GL and let fall 0/ 

 perpendicular to CD; — then since the angle CO/ = GPL 

 and the including sides CO, /O, just half the sides PG, PL, 

 the triangles are similar, and the angle PLG = a right angle. 

 Consequently the required locus of L is a circle whose radius 

 PS = CO." 



" From the centres 



Fig. 2 



