Strong’s Problems. ‘OTF 
Const. Let (Fig. 9. pi. 2.) ALA’, BMB’, CNC’, be 
the given circles, of which ALA’, BMB’, are equal, and 
CNC’ is less than the other two. Let G, H be the centres 
of ALA’, BMB’, From G, H, as centres, with radius =ra- 
dius of ALA—radius CNC, describe the circles, DD’, EE’. 
Through F the centre of the cirele CNC’, describe the cir- 
ele FDE, (Prob. IX.) touching DD’, EE’ in D, E, of 
which circle, let O be the centre. Join CD, and it will pass 
through G. Then from O as centre, with radius =OD+ 
CF (=radius of the circle CNC’) describe the circle ABC, 
which shall be the circle required. 
Demonstration. Fer join ODA. Now because ODA= 
radius of the circle ABC’ (=OD+DA, or FC) of which 
O is the centre, A is in the circumference of ABC. And 
because ODA passes through G, and GD+DA=radius of 
the circle ALA’, A is in the circumference of ALA’. Hence 
LA’, ABC meet in A. And they likewise touch in A. 
For if Az be drawn at right angles to ADO, A will be a 
tangent to both circles in the same point A. Whence the 
circles must likewise touch in that point. In like manner 
itmay be proved that ABC, BMB’, ABC, CNC’, touch 
each other respectively at the points B,C. ABC is there- 
fore the circle required. ; 
Now by using OD’— AD’ for OD+AD, and A’, B’, &e. 
for A, B, &c. the demonstration is the same when none of 
the given circles is comprehended. 
When the touching circle comprehends both the equal 
eircles, and touches the smaller one externally ; or compre- 
hends the smaller circle, and touches the equal circles ex- 
ternally. 
Construction. Let (Fig. 10. pl. 2.) Gy, Fx, Iz, be the 
given circles of which Gy=F vr. Let A, B, C be the cen- 
__ Demonstration. For join ; , which will pass 
through ¢ centres A, B. Now (by Const.) D is in the 
