274 Strong’s Problems. 
demonstration answers = trap 16, in which the larger cir~ 
cle is alone comprehe 
6. When the ester ‘circle comprehends one of the 
equal circles, together with the larger, or one of the equal 
circles alone. 
Const. Let (Fig. 1. pl. 3 ete HPN, MGQ, DER be the 
given circles of which DER=MGQ. Let A, B, C, be the 
centres of these circles respectively : From A with radius 
==radius of the circle HPN + radius of the circle DER de- 
scribe the circle IK : and from B with radius=two radii of 
the circle MGQ describe the circle FL. Through C de- 
scribe the circle CFI touching IK, FL in the pots I, F ; 
of which circle let O be the centre. Decreasing the redius 
by a line=radius of the circle DER describe the circle HE 
G which shall be the circle required. 
Demonstrations. For join OC. Let OC cut the izle 
DER in the point E. Now: because OE=OC—CE, 
. = circumference of the circle HEG. — If therefore ( Fi, ie. 
3.) as in former cases, a perpendicular be erected at 
a it will touch both circles at that point. Therefore the 
circles DER, HEG touch each other in the point E. In 
like manner it may be shown that the circles HPN, HEG; 
GQM, HEG respectively oe at the points H, G. There- 
fore HEG is the circle abe ds 
using OC +CE for OC-_CE this demonstration 1s 
applicable to Fig. 2, in which the touching circle compre- 
hends one of the smaller circles and touches the other, 0% 
gether with the larger circle externally. 
Case If1. When all the circles are unequal, 
1. When the touching circle cuthighenids all or none of 
the given circles. 
. Let (Fig. 3. pl. 3.) AL, HN, MF be the given 
penn gd which F'M is the least. Let B, K, E, be three cen- 
of the circle AL—radius of the circle FM. From K de- 
scribe the circle IP, whose radius==radius of the circle NH 
—radius of the circle FM. Through E describe the circle 
EDI touching DG, PI in D, I. (Prob. V.) Let O be the 
centre of this cirele. Frou O, with radius—radius of the 
circle DEI+radius of the civelé FM, describe the cirele 
AFH, which shall be the circle requ uired. 
