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 
x, y, describe the circle cBy touching BB’g in B, (Prob. 
M11.) 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. 
Demonstration. For join OBA which will pass through 
G the centre of the circle BB’g. Now because OA=O 
B+BA and GA=GB-+BA, A is in the circumferences 
ALA’, ADE. Hence if a line be drawn at right angles to 
~ OA at the point A it will be a tangent to both circles at the 
same point A. Hence the cicles ALA’, ADE touch each 
other at the point A. In like manner it may be proved, 
that the circles DND’, ADE’, EME’, EAD, touch respec- 
tively at D, E. 
By joining O’A’B’, and using A’B’, &c. for A, B, &c. and 
O'B’—O’'A’ for OB+ AO the demonstration is the same 
when the circles are none of them comprehended. 
5. When 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 
circles of which Hy=Iv. 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- 
cles 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.) Q as centre with radius=radius 
of the circle AED—radius of the circle Hy describe the 
circle LHFI, which shall be the circle 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 
LL.MH is in the circumference of the circle HFJ. 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 Iv, HFI; ML, HFL 
touch respectively at the points I, LL. HEI is therefore the 
circle required. By using EQ+EH for OEF—EH. this 
