152 MR. T. T. WILKINSON ON 



at P. Demit B q perpendicular to P D, and on D F apply 

 B K and B L = B g-. Join P K and P L, which are the lines 

 to be found. On P L demit the perpendicular B m, and take 

 S' such that B D' : B w* : : S* : 2 O S . S F, and S» is the 

 space." (Math. Comp., dues. 606.) 



Mr. Simpson remarks that his demonstration points out 

 solutions to the following Problem and Porism : — 



(1). " Three circles given in magnitude and position, whose 

 centres are in the same right line, it is required to find a point 

 in one of them, such that tangents being drawn to the other 

 two their rectangle may be given." 



(2). "Two circles, one included within the other, being 

 given in magnitude and position, two right lines also given by 

 position, may be found, such, that if from any point in the 

 lesser circle perpendiculars be demitted on the lines so found, 

 and through the same point an ordinate to the greater, the 

 sum of the squares of the perpendiculars will have a constant 

 ratio to the square of the ordinate." 



PORISM. 



Proposed hy Mr. John Kay. . 



" A and B are given points in the arc of a given semicircle 

 to which tangents are drawn to intersect in D ; now a right 

 line may be found, such, that if perpendiculars be demitted 

 from any point O in the arc A B, upon the lines AD, B D, 

 the sum of the squares of those perpendiculars with r times 

 the square of the perpendicular from the point O upon the 

 line to be found, will be a constant magnitude ; r being a 

 constant quantity whose value is to be found from the data." 



Porismatic Construction. By the Proposer. 



" Generalise the proposition by taking the lines DA, D B, 

 equidistant from C, the centre of the whole circle. Draw z x 

 parallel to D A, to meet D Z C in the point z of the 

 periphery ; draw z m perpendicular to D s C P, and from P, 



