560 
[383 
383. 
PROBLEMS AND SOLUTIONS. 
[From the Mathematical Questions ivith their Solutions from the Educational Times, 
vols. I. to iv., 1863 to 1865.] 
[Vol. i. (June 1863 to June 1864), pp. 18, 19.] 
1373. (By T. T. Wilkinson, F.R.A.S.)—Given a circle (G) and any point A, either 
within or without the circle: through A draw BAD cutting the circle in B, D. 
Then it is required to find another point E, such that, if LEM be drawn cutting 
the circle in L, M, we may always have AE 2 = LE. EM ± BA . AD. 
Solution by Professor Cayley. 
Consider a circle centre 0 and radius OA, and in relation thereto a point M 
either outside or inside the circle, and suppose that 
(OA) 2 — (OM) 2 , or the “squared inner potency” of M is denoted by □ i.M, 
and 
(OM) 2 — (OA) 2 , or the “squared outer potency” of M is denoted by □ o.M, 
so that, for an outside point, □o.ilf, = — CU.M, is the square of the tangential 
distance of M from the circle; and, for an inside point, Qi.M, =—□ o.M, is the 
square of the shortest semi-chord through M. 
Suppose now that M is a given point; the proposed question is in effect to find 
the locus of a point P such that + □ o . P ±Uo .M = (MP) 2 ; but we have thus in 
reality four different questions according as the signs are assumed to be + +, H—, —f-, 
or ; the case + + , or when Do. P + Do . M = (MP) 2 , is perhaps the most inter 
esting.