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PROFESSOR J. CASEY ON A NEW 
then if the diameter of the generating circle be 2 a, we have, from art. 97, 
2 ao 
n= — n ■ 
q — 2 a 
Now let the angle IBP' be denoted by 4/ and IP' by b, and we have 
sin 20 : sin ^ : : BP' : b. 
But 
hence 
Again, we have 
BP'=BK+KP'=£(c cos \|/+ if c=“ ; 
20=sin _1 -|sin \|/(c cos + A\J/) } . 
a : <r' : : % : g — la; 
a' ~2a6, 
o—a sin -1 -jsin \J/(c cos \|/ + A\f/) j- (233) 
And this is the equation of the evolute of the extern epicycloid. 
115. From the diagram we have <J>=<p + \f/ ; 
6?<E> dq> 
da' d<r'‘d<r' 
Now let a consecutive position of the line P'B intersect P'B in the point Q, and the 
arc OB in the point B' ; then if I) denote the diameter of the circle described about the 
intinitesimal triangle BB'Q, we have 
d&_ 1 
d<r'— D’ 
and from equation (233) we get 
__ A(4Q . 
d<r' a(c cos 4' + A(\J/)) ’ 
. 1 _ 1 A(49 
D g — 2a~t~a(ccos\{/ + A(\J/))’ 
Hence 
a (40 
(234) 
Dcos4' (g — 2a) cos4/~o cos ^(ccos^-t- A49’ 
If the polar of the point P' with respect to the generating circle meet BP' in N, we 
have, see art. 86, equation (207), 
a cos \|/(c cos \J/ + A\J/) 
BN: 
a (40 
Hence, from equation (234), if CK' be perpendicular to BP', we have 
