90 Improvements in Fundamental Ideas 



on the given lines are specified, then there is but one answer- 

 able point A, one circle AIB, two points E real or unreal, 

 and therefore but two solutions. But if m.n be not restricted 

 as to sign, there are then two points A, two circles AIB, and 

 therefore four answerable points E and .\ four solutions. 

 Moreover, since the two points A must be on opposite sides of 

 MM, it follows that one of the circles AIB will always cut 

 MM in two real points E. 



Limiting values for the angular magnitude c 6 right.' 



To find the limiting values of the angle ' 9 right ' when the 

 rest of the data is unchangeable, we may proceed as follows: 

 — Thus, looking on the triangle DEO, we see the angle 

 DO right to E is constant ; and it is evident that when OE 

 right to D or OB right to C is at its limit, then must ED 

 right to O or EA right to B be at its limit. But this last 

 angle is evidently at a limit when the circle BAE touches 

 MM. Hence it is evident that by describing the two circles 

 through A and B which touch MM, and putting i and i for 

 the other points in which they cut circle ACH, then will the 

 angles iB right to C, and iB right to C, be the required limits. 

 And it is, moftover, evident that according as the magnitude 

 of any angle ' 6 right ' is not comprehended between these 

 limits, or equals one of them, or is comprehended between 

 them, so accordingly will the corresponding circle AIB cut 

 MM in two imaginary, in two real and coincident, or in two 

 real and distinct points E. 



Limiting values for m.n. 



The limiting values of m.n, the rest of the data being un- 

 changeable, may be found as follows : — 



It is evident the point A varies on PA according as the 

 values of m.n vary, and that when m.n is at its limits, the 

 points A will be in limiting positions on PA. Now it is, 

 moreover, evident that the other point K in which circle 

 AEB cuts PA is fixed and given, no matter how A may vary 

 on PA (because angle KA right to B = EA or ED right to 

 B or O, and that the angles OB right to C and DC right to 

 A, or which is the same thing, because the angles OE right 

 to D and DO right to E are given in magnitudes). But the 

 points A are in their limiting positions when the circles 

 AEBK are in limiting positions or touching MM. Hence 

 putting a and a for the other two points in which the circles 

 through B and K to touch MM again cut PA, we shall have 



