600 
PROBLEMS AND SOLUTIONS. 
[485 
[Yol. xl, January to June, 1869, pp. 33—38.] 
2718. (Proposed by Professor Cayley.)—Find in piano the locus of a point P, 
such that from it two given points A, G, and two given points B, D, subtend equal 
angles. 
2757. (Proposed by Professor Cayley.)—If 
a?o 2 + 2/o 2 = 1, 
aY + 2/i 2 = 1, 
show that each of the equations 
a 2 {x - xf + b- (y - yf 
(xx 0 + yy 0 - l) 2 
a 2 (x - x 0 ) 2 + b 2 (y - ,y 0 ) 2 
(xy 0 - x 0 y) 2 - (x - x o y -(y- y 0 ) 2 (xy 1 - x x yf -{x- xf 2 -(y- y x f ’ 
represents the right line L = 0 and a cubic curve. 
1819. (Proposed by C. Taylor, M.A.)—From two fixed points on a given conic 
pairs of tangents are drawn to a variable confocal conic, and with the fixed points 
as foci a conic is described passing through any one of the four points of intersection. 
Show that its tangent or normal at that point passes through a fixed point. 
% , y , 1 = L ; 
«o, 2/o> 1 
> Vi> 1 
a 2 (x - xf 2 + b 2 (y- yf 2 
(1) 
(xx, + yy Y -l) 2 
a 2 (x - x x y + b 2 (y- I/,) 2 
Solution of the above Problems by Professor Cayley. 
1. It is easy to see that drawing through the points A, C a circle, and through 
B, D a circle, such that the radii of the two circles are proportional to the lengths 
AG, BD, then that the required locus is that of the intersections of the two variable 
circles. 
Take AG =21, MO perpendicular to it at its middle point M, and =p\ a, b the 
coordinates of M, and \ the inclination of p to the axis of x; then 
coordinates of 0 are a + p cos X, b + p sin X, 
coordinates of A, G are a ± l sin A,, b + l cos X, 
and hence the equation of a circle, centre 0 and passing through A, G, is 
(x — a —p cos X) 2 + (y — b —p sin A) 2 = P + p 2 ; 
or, what is the same thing, 
(x — a)- + {y — b) 2 — l 2 =2p [{x — a) cos A + (y — b) sin A].