CONCERNING TERRESTRIAL MAGNETISM. 
81 
Lemma V. Theorem. — If lines be inflected from two points in a given straight line 
to two given points, the pair which is more remote from the point at which the ratio is 
a minimum will have a greater ratio than those inflected from the point which is nearer \ 
And when their order is changed, the ratio will be less. 
Let T, U be the points, and suppose T the point nearer to the given line N S to 
he the point from which the antecedent line is drawn. Let P be the point found in 
the last lemma, and N nearer to P than N' is ; then the ratio N' T : N' U is greater 
than the ratio N T : N U. (Plate XI. fig. 4.) 
For divide the line T U harmonically in the points C and D in the ratio NT : N U, 
and describe the circle on C D, passing through N (Lemma 1.). 
Then since N is nearer to P than N' is, the point N' falls without the circle DNC; 
and hence (Lemma II.) the ratio N T : N U is less than N' T : N' U. — Q. E. D. 
Lemma VI. Problem. — From two given points T, U within a circle to inflect lines 
to the circumference, so that they shall have the greatest or the least ratio possible. 
(Plate XI. fig. 5.) 
Suppose the points to be found at P and P' ; and draw the tangent P H meeting the 
perpendicular M H from M, the middle of the line joining the given points. 
Then from the reasoning in Lemma II., the circle of ratios will touch the given 
circle in P and divide T U harmonically in C and D ; and from Lemma IV.* we learn 
that H P = H T. Hence the problem is reduced to finding a point H in the line M H, 
from which tangents being drawn to the given circle RPE they will be equal to 
H T or H U. 
Though A the centre of the given circle draw A N parallel to T U ; then it is per- 
pendicular to H M or E K, and E K is bisected in N. Then 
HP 2 = KH.HE = HT 2 = TI# + MH 2 
= T M 2 + (N H - N M) 2 
= T M 2 + N H 2 2 IT N . N M + N M 2 . 
Hence 2 H N . N M = T M 2 + N E 2 + K II . II E - K H . II E + N M 2 
= T M 2 + M N 2 + N E 2 
= T N 2 + N E 2 . 
Whence we have to form a rectangle whose area is T N 2 + N E 2 , and one of whose 
sides is 2 N M ; and the other side of the rectangle is the distance of H from N. 
Construction. — With centre N and distance NT describe a circle cutting AN in Q. 
Join E Q, and make N G = E Q, and NB = 2N M. Join B G, and draw G PI per- 
pendicular to it, cutting M II in H. Then with centre H and distance H T or H U, 
describe a circle cutting the given circle in P and P', and these will be the points 
required. 
* Since DPT touches the line H P and divides T U harmonically. 
MDCCCXXXVI. 
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