PROBLEMS AND SOLUTIONS. 
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485] 
If 2m, q, c, d, p refer in like manner to the points B, D, then the equation of a 
circle, centre say Q, and passing through B, D, is 
(x — c) 2 + (y — d) 2 — m 2 = 2q \{x — c) cos p+{y — d) sin /a] ; 
and the condition as to the radii is l 2 + p 2 : m 2 + q 2 = l 2 : m 2 , that is, p 2 : q 2 = l 2 : m 2 , 
or p : q= ±1 : m. And we thus have for the equation of the required locus 
(x — a) 2 + (y — by — l 2 _ ^ l (x — c) 2 + (y — d) 2 — m 2 
(x — a) cos \ + (y— bj sin A ~ m {x — c) cos p + (y — d) sin p ’ 
viz. the locus is composed of two cubics, which are at once seen to be circular cubics. 
One of these will however belong (at least for some positions of the four points) to 
the case of the subtended angles being equal, the other to that of the subtended 
angles being supplementary; and we may say that the required locus is a circular 
cubic. 
2. If two of the points coincide, suppose G, D at T\ then, taking T as the 
origin, we may write 
a = l sin A, b = —l cos A, 
c = — m sin p, d= m cos p, 
B 
and the equation becomes 
x 2 + y 2 + 21 (x sin A — y cos A) _ + l x 2 + y 2 + 2m (x sin p — y cos p) 
x cos A + y sin A ~ m x cos p + y sin p 
viz. this is 
(x 2 + y 2 ) [m (x cos p + y sin p) + l (x cos A + y sin A)] — 2Im {(x sin A — y cos A) (x cos p + y sin p) 
+ (x sin p — y cos p) (x cos A + y sin A)} = 0. 
Taking the lower signs, the term in { } is (x 2 + y 2 ) sin (A — p), and the equation is 
(af + y 2 ) [m (x cos p + y sin p) + l (x cos A + y sin A) — 2Im sin (A — p)\ = 0, 
viz. this is x 2 + y 2 = 0, and a line which is readily seen to be the line AB; and in 
fact from any point whatever of this line the points A, T and the points B, T 
subtend supplementary angles. 
C. VII. 
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