and Loci of Apollonius, ^c. 59 



: : AN : : fc'; on AB and AC make Gp and llq equals to y 

 and z, and describe the cireles AGH, Apq; tlirougli the point 

 of intersection I of these two circles draAv IE perpendicular 

 to AI to cut the circle AGH again in R; find the point i in 

 IR so that AI : li : : AM : e'b', and draAV ii' parallel to AG 

 to cut AI in i' ; with R as centre and radius equal ti' describe 

 a circle, and from either point S in Avhich it cuts the circle 

 AGH draw AS to cut IR in O. 



Then will O be the centre of a required circle. 



NOTES. 



This method of solution holds good in all cases in which 

 circle A is not supposed infinite, or — which amounts to the 

 same thing — it holds good in all cases but those in which we 

 suppose the three given circles replaced by straight lines. 



If we suppose circle C infinite, then it is evident AC is 

 parallel to OC and perpendicular to tangible portion of the 

 infinite circumference; moreover since the ratio of AN to fc 

 is then one of equality, so will that of AO to Q.H be one of 

 equality, and .'. QH will be = Of, and the point H in the 

 perpendicular from A on the known portion of the infinite 

 circumference is knovv^^ as it is at a distance = radius of 

 circle A therefrom. 



It may also be observed that the line lOR is identical Aidth 

 the line lOR of the seventh solution. 



NINTH SOLUTION. 



(See Plate.) 



To describe a circle to touch three given circles A, B, C. 



ANALYSIS. 



Let O be the centre of tlie required circle, and let D, E, F 

 be its points of contact with the given circles A, B, C. 



If in OA we suppose OM taken equal OB and so that OB 

 and OM have like directions in respect to the directions OE 

 and OD, then DM = EB, and it is evident AM is of known 

 magnitude. 



And since OB is equal OM and that 2.0A.OM = OA^ + 

 OM2 — AM2 . 



.•• 2.0A.0M = 0A2 + 0B2 — AM^ both in sign and 

 magnitude. But if G be the point in which the circle haxnng 



