396] 
TRIANGLE INSCRIBED IN A CIRCLE. 
75 
16 
is 
ly 
ill 
I proceed to find the tangential equation of the envelope. Representing the 
equation of the line by 
& + r ny + & = 0, 
we have 
£ : ^ : £ a ^ ^ : ^ — v) (fi — X) : - v (v — A) (i/ — f), 
or, what is the same thing, 
£ : V : £ = - 
a 
A 
fi — v 
where as before 
and eliminating A, i/, we find 
. 1 P 
b v — A 
^ ^ . V 
- + i + - = 0, 
a b c 
1 ^ 
c A — /a ’ 
at 
a* fa - D 2 + k? (C ■- I) 2 + (f - T;? = 0. 
:W 
;h 
e, 
)f 
'a; 
e 
11 
In fact we find at once 
aÇ (y~Çy ■ br](Ç- I) 2 : c£(£ - 77? = fa — i>) A /a (A — ¿i,) — ^ v (1/ — A)| 
: fa — A) fi 1/ fa — 1/ ) — 1 A (A — fi) 
: (A - fi) v jï Afa — A) — i/i fa - v )j , 
and the sum of the three expressions on the right-hand side is 
= “ fa “ v) ( v ~ A) (A — f) Qj + ^ = 0, 
which verifies the result just obtained. 
The tangential equation of the envelope is thus 
(77 - O 2 + by fa- & + fa fa - 77? = 0, 
or the envelope is a curve of the third class having as a double tangent the line 
£ = 77=£ that is the line infinity; in fact for these values the equation & + v y + & = 0 
becomes x -1- y -1- z — 0, which is the equation of the line infinity. The curve is therefore 
a curve of the fourth order, the equation of which is 
—— recip. [cfa (77 — £) 2 + brj fa— £) 2 + fafa — 77)-] — 0, 
{x + y + zf 
10—2