Strong’s Problems. 273 
LA’ the other whose centre is G. From G as centre with 
radius = radius of the circle ALA‘—radius of the circle 
EME, describe the circle BB’g. Then through the points 
ne describe the circle rBy touching BB’g in B, (Prob. 
LI.) of which circle let O be the centre ; increasing the ra- 
dius by a line = radius of the circle EME, describe the 
circle ADE, which shall be the circle required. 
tively at D, E. 
By joining O'A’B’, and using A’B’, &c. for A, B, &c. ard 
0'B’—O'A’ for OB4+AO the demonstration is the same 
when the circles are none of them comprehended. 
hen the touching circle comprehends both of the 
equal circles; and touches the other externally, or com- 
prehends the larger and touches the other externally. 
Const. Let (Fig. 15. pl. 2.) ML, xl, Hy be the given 
eircles of which Hy=Ia. Let A, B, C be the centres of 
these circles. From the centres C, B with radius=radius 
of the circle Hy+radius of the circle ML describe the cir- 
eles EN, DP. Then through A describe the circle AED 
touching EN, DP in E, D, of which let O be the centre. 
From (Fig. 16. pl. 2.) O as centre with radius=radius. 
of the circle AED—radius of the circle Hy describe the 
circle LHF I, which shall be the cirele required. 
Demonstration. For join OCE which as before shall 
pass through the centre C. Let it cut the circle Hy in H. 
Then because OH==OE—HE or the radius of the circle 
LMH is in the circumference of the circle HFI. 'There- 
fore if a line be drawn perpendicular to OC at the point it 
will be a tangent to both circles Hy, HIF at the point H. 
The circles therefore touch at the point H. In like manner 
it may be shown that the circles Iz, HFI; ML, HFL 
touch respectively at the points I, L. HF 'I is therefore the 
circle required. By using EO+EH for OR—EH this 
