D4 2: Strong’s Problems. 
circumferences DCE, gD. Therefore since OG—=OD— 
GD, or OD—CI, and AG=AD—CI, G is in the circum- 
ferences of GIE, Gy; and if at Ga line be drawn at right an- 
gles to OG, it will touch both circles at the same point G. 
‘Therefore they touch each other at the point G. In like 
manner Iz, GIE ; Ev, GIE touch respectively in I and E. 
Wherefore GIE is the circle required. Now by using 
OD+GD for OD—GD, the demonstration is the same in 
Fig. 11. 
5. When the touching circle circumscribes one of the 
equal circles alone, or one of them together with the small- 
er one. 
Construction. Let (Fig. 12. pl. 2.) Eg, Hz, Ba, be the 
given circles, of which A, F, C, are the centres. From F 
as centre with radius—radius circle Hz—radius of the cir- 
cle Bx (i. e. FH—BA) describe the circle Gy. And from 
C as centre with radius—radius of the circle Hz or Eg+ 
vadius of the circle Bx describe the circle Dn. Then de- 
scribe through A the centre of the circle Bz, the circle AG 
D, (Prob. V.) touching (Fig. 13. pl. 2.) Dn, Gy in D, G; 
of which circle let O be the centre. From O with radius 
= radius of the circle AGD—radius of the circle Br de- 
scribe the circle HEB, which shall be the circle required. 
Demonstration. For jon CHG, OCED, OBA. The 
line OHE will pass through G, the point of contact of the 
circles Gy, GAD. Now because GH=BA and OG—B 
A==OH, the point H is in the circumferences of the cir- 
cles Hz, HEB. And if a line be drawn at right angles to 
OG atthe point H, the circles Hz, EHB will touch it at the 
same point H. Therefore they touch each other at that — 
point. In lke manner it may be shown that the circles Ba, 
BHE, Eg, EHB touch each other respectively at the points 
B, E. Therefore EHB is the circle required. 
If instead of OG—BA, OG+BA be used, this demon- 
stration is applicable to Fig. 13, in which one of the equal 
circles, and the smaller one are comprehended by the 
touching circle. 
4. When the two equal circles are less than the other, 
and when the touching circle comprehends all or none of 
the given circles. 
Construction. Let (Fig. 14. pi. 2.) DND’, EME be 
the two equal circles, of which 2, y, are the centres, and A 
