demonstrated by Mr. Lubbock in Phil, Mag. for August, 33 



be found, but, if I recollect rightly, there is more than one 

 solution of it. The most simple way of considering the ques- 

 tion is undoubtedly by geometry, as the following demonstra- 

 tion will testify ._ 



Let A P, A Q, B C be 

 the three tangents intersect- 

 ing in ABC. Draw the 

 diameter F S T, meeting 

 A Q and A P in S and T. 

 Join FP, FQ. Then, by 

 a known property of the pa- 

 rabola, PFX = 2PTF, 

 QFX = 2QSF. Also 

 BAC = PTF + QSF. 

 Therefore PFQ=2BAC. 

 Q Again, join F B, F C, F R. 

 By a property of the para- 

 bola P F B = B F R and 

 Q F C = C F R. There- 

 X fore the salient angle P F Q 



= 2 B F C, and all the angles round F or 4 right angles 

 = 2 (B A C + B F C), or B A C + B F C = 2 right angles, 

 and therefore a circle can be described round A B C F. 



But if an analytical proof of the theorem be required, the 

 following, by Mr. Greatheed, of Trinity College, deserves 

 attention from its elegance and brevity. 



Let the curve be referred to one of the tangents and the 

 diameter through the point of contact : then its equation is 

 y2 = 4< Wi .r ; and the equations to the two other tangents, 

 whose points of contact are x^ y^ x<^ y,^^ are 



y yi = 2 7»i (cT + jTi), ^^2= 2 m, (07 + 0:2). 

 These tangents will cut the axis of y at distances -^, 

 and the coordinates of their point of intersection are 



Ml. 

 2 



y^ +^2 

 4. m, ' 2 



y\y<i 



If t be the angle between the axes, the equation to a circle is 



x'^ -\- 2 xy cos ^ -^ y'^ + ax + by + c = 0. 

 When ;r = 0, ^- + 6j/ + c = 0. 



. Therefore 



But the roots of this must be ■^, 



ThirdSeries, Vol. 10. No. 58. Jan. 1837. 



F 



