ON THE CAUSTIC BY KEFLECTION AT A CIBCLE. 
[From the Cambridge and Dublin Mathematical Journal, vol. II. (1847), pp. 128—180.] 
The following solution of the problem is that given by M. de St-Laurent (Annales 
de Gergonne, t. xvn. [1826] pp. 128—134); the process of elimination is somewhat 
different. 
The centre of the circle being taken for the origin, let k be its radius; a, b the 
coordinates of the luminous point; y those of the point at which the reflection takes 
place; x, y those of any point in the reflected ray: we have in the first place 
? + y 2 = k 2 (1). 
There is no difficulty in finding the equation of the reflected ray 1 ; this is 
(&£ - arj) (ijx + yy - k 2 ) + (y£ - xrj) (af + by - Jc 2 ) = 0 (2), 
1 To do this in the simplest way, write 
p- = (f - *) 2 + (v~ V?> o- 2 =(I - a) 2 + (v~ W, 
then, by the condition of reflection, 
p + (r=min., 
p, <r being considered as functions of the variables £, r], which are connected by the equation (1). Hence 
izf + L^+^o, 
or, eliminating X, 
lLi' + tf + x , =0; 
p a 
■qx-jy V a -&_Q 
P a 
whence 
This may be written 
(yx - £y) 2 [(I - a) 2 + ( v -b) i ] = (7ja - £i) 2 [(| - x) 2 + (y- y) 2 ]. 
{{yx-Zy) (l-a)-(v«-^) (£-*)} l(yx-£y) (£-a) + (ya-(b) (£-&)] 
+ {(yx-%y) {y-b)-{ya-£b) (y — y)} [(yx-£y) {ti-b) + (yia-£b) (y-y)] = 0 ; 
the factors in -[ } reduce themselves respectively to £P and rjP, where P=£ [b -y) - r] (a- x) + ay -bx ; omitting 
the factor P, (which equated to zero, is the equation of the line through (a, b) and (£, yj),) and replacing 
f (£ -a)+ 7j (r)- b) and £ (£ - x) + rj (77 - y) by k 2 -a%-b-q and k 2 -£x-7]ij, respectively, we have the equation given 
above. 
c. 35