Strong’s Problems. 267 
segment RBH, and the angle THO=RHM==angle in the - 
segment HLKI, But the angle in the serment RBH= 
angle in the segment IDEP. Therefore the angle in the 
segment IDEP=angle in the segment HLKI—angle HIQ, 
or NIP. Whence the angle in the segment IDEP=NIP; 
wherefore NQ touches DEPI in the point I. Conse- 
quently, HLKI touches IDEP in I. Now (by Const.) 
HLKI touches RBH, and passes through L. Wherefore 
it is the circle required. 
ase IT. When the two circles are unequal, and the cir- 
cle which passes through the given point circumscribes 
them. 
Const. Let (Fig. 2. pl. 2.) as in Case i. the point be 
L, and the circles HBC, DGE. Draw the tangent FG, 
and extend it till it meet zy, produced in A. Let vy pro- 
duced cut the given circles in C, B, E, D. Through L, 
and C, D, the remote points in which ry cuts the given cir- 
cles, describe (Prob. i.) the circle LCD. Suppose AL 
produced to meet the circumference of this circle in K. 
Through the points, K, L, describe (Prob. v.) a circle 
touching HBC, in H. Then shall this be the circle. re- 
quired. 
~The points AI’, Al being joined as in case I. and the tan- 
gents MO, NQ being ‘diawn, the demonstration employed 
in case 11, is applicable to this. 
Case I. When the touching circle circumscribes one 
of the given circles, and touches the other externally. 
ences of the circles in B, C, D, E. Draw FG, a tangent to 
the circles in the points F, G. Let this cut the line, C 
nA. Join AL. Through L, C, D, describe the circle L 
CD. Suppose AL extended cuts this circle in K. Through 
L, K, describe the circle LKH touching BHC in H. Join 
AH, and let AH produced cut DGE in L, and HKL in 
['. Draw, as in cases I and 11, MHO touching the circles 
HKL, BHC, in H, and NQ touching HKVL in V. 
“Now by applying the Demonstration in case I, the circle 
HKI'L, as in former cases will be found to answer the con- 
ditions of the problem. 
