PROBLEMS AND SOLUTIONS. 
568 
[705 
or 
that is, 
k 2 — a 2 e 2 = 3 (a 2 — k 2 ), 
3 a 2 — 4& 2 + a 2 e 2 = 0 ; 
viz. this is the condition for the existence of the three circles U, V, W, each touching 
the two others, and also the circles R, S. 
The circle R lies inside the circle S, and the tangential distance is thus 
imaginary; but defining it by the equation 
squared tangential dist. = squared dist. of centres — squared sum of radii, 
the squared tangential distance is 
= 4a 2 e 2 — 4a 2 . 
But if the circles were brought into contact, the distance of the centres would be 
2k, and the value of the squared tangential distance = 4k 2 — 4a 2 ; hence, if this be 
= one-fourth of the former value, we have 
that is, 
4 (k 2 — a 2 ) = a 2 e 2 — a 2 , 
3a 2 - 4>k 2 + a 2 e 2 = 0, 
the same condition as above. The solution might easily be varied in such wise that 
the circles R, S should be external to each other, and therefore the tangential distance 
real; but the case here considered, where the locus of the centres of the circles 
U, V, W is an ellipse, is the more convenient, and may be regarded as the standard 
case. 
[Yol. xiv., p. 19.] 
3144. (Proposed by Professor Cayley.)—If the extremities A, A' of a given line 
AA' describe given lines respectively, show that there is a point rigidly connected 
with A A' which describes a circle. 
[Yol. xiv., pp. 67, 68.] 
3120. (Proposed by Professor Cayley.)—To find the equation of the Jacobian of 
the quadric surfaces through the six points 
(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), (1, 1, 1, 1), (a, /3, 7 , S). 
Writing for shortness 
Solution by the Proposer.