508 Analytical Solution of the Problem of Tactions. 



A=/3y'-^'7+/3'7"-W + /3"7-/8y', 



B = yci' — r/a + 7'a" — yy + y"u — yu", 



C = «^' - u'^ + «'/3" - u"/3' + a" 13 - «/3", 



V = u^'y" - a^"y' + ci'l3"y - u'^y" + u"l3y' - cJ'^'y, 

 values which, it will be observed, give 



Aa +B^ +C7 =V, 



A«' +B;8'+Cy=V. 



A«" + By8" + C7"=V. 

 Then the equation of the axis of similitude is found to be 



Aa,' + B?/— V=0. 



Whence also the equation of the perpendicular let fall from the 

 radical centre upon the axis of similitude is 



Bx — ky = Q. 

 It should therefore be possible to find two circles having their 

 centres on the last-mentioned line and touching the three given 

 cn-cles. Take A^, B^ as the coordinates of the centre of one 

 of the two circles, and let r be its radius ; the conditions of tan- 

 gency are 



r= \/{kd-uY^ (B^-/3)2±7, 



= V[ke-u<f+{m-^'f±y', 



= V(A^-a")2 + (B^-/3")2±7", 

 where the sign + has the same value in each expression. We 

 have consequently 



(r + 7)2=(A^-a)2+(B^-/8)2; 



or, observing that Aa + B/S = V — C7, and reducing, 



?-2-(A2 + B2)^H2V^-l = 27(±?- + C^). 



Forming the two analogous equations, the three equations will 

 be satisfied if only 



,.2_(A2 + B2)^2 + 2v^-l=0, 



±r + Gd =0. 



Eliminating r, we have 



(A2 + B2-C2)^2_2v^+1 = 0, 

 which gives for 6 the two values 



(A2 + B2-C2)^=V± /V2-(A2 + B2-C2); 

 and then r is determined linearly by the equation 



