98 PHILOSOPHICAL TRANSACTIONS. [aNNO 1673. 



circle pass through the three points A^E, B; and AO be produced to its cir- 

 cumference at D. Then if the point O fall between E and B, the right line 

 A O is to be produced towards O ; but if beyond B, and the rectangle EOBbe 

 greater than the square of AO, it must be produced towards A ; but if the 

 rectangle be less than that square, the circle will cut the right line AO in the 

 very point D. Then drawing AX parallel to EB, cutting the given circle in 

 N, make the rectangle DAO to the square of AN, as -l-AX is to AH; which 

 must be taken towards X, if O fall between E and B, or if the rectangle EOB 

 be less than the square of OA ; but on the contrary side, if it be greater. Now 

 put OQ equal to AH, in the direction EB in the first or second case, but to- 

 wards E in the third. Then let X A, NA, HK be proportionals, and all taken 

 in every case towards X ; and let AO be cut in V, so as that K A may be to 

 A V as AD to AX; join KV, and produce it to meet the right line QM, 

 parallel to OA, indefinitely produced, in the point L. Then in every caseKL 

 and QL will be the asymptotes of the hyperbola, which drawn through the 

 point O will satisfy the question. Yet with this difference, that in the first or 

 second case, the hyperbola passing through O will answer for the convex specu- 

 lum, and the opposite section for the concave one ; but, on the contrary, in 

 the third case, the hyperbola through O will answer the concave speculum, and 

 its opposite the convex. And thus it will be when the point V falls between A 

 and O ; for if it should fall beyond O, one hyperbola alone, drawn between the 

 same lines QL and KL, will serve for both the convex and concave specula. 

 But if V should fall in the point O, then it would be a plane problem, and the 

 right lines LQ and LK would resolve it. Whence it appears that this problem 

 admits of infinite cases, which may be solved by a plane locus: and hence those 

 are excusable who have thought the problem might be solved universally by the 

 same locus, because thus sometimes the calculus has been successful. For 

 there can be given no position of the three points A, E, B, (still reserving the 

 case of equality of the rectangle EOB and the square of OA for future con- 

 sideration), which admits not of some circle to be described from the centre A, 

 at the circumference of which the problem may be solved by a plane locus. As 

 to the radius of this circle, if worth while, it might be thus found : In the first 

 and second case of the above construction, make as AX^ -j- double the rect- 

 angle OAD to 2AD^ so AO^ to AN^ then AN will be the radius sought. 

 But in the third case, there is to be made as AX^ — the double rectangle OAD 

 CO !2AD% so is AO' to AN^ 



There now remains to be constructed the other case, viz. when the rectangle 

 EOB is equal to AO^, or in which the circle described through the points 

 A. B, E, touches the right line AO. And here M. Huygens has rightly ob- 



