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Strong’s Problems. O05 
to 
Demonstration. For jom ODA which will pass through 
the centre B. Now because O0A=OD+EM and BA—=D 
B+EM, (if M be the point in which a line joining O, E, 
cuts the circle F'M) A is in the circumferences of AL, AFH. 
And if at A a perpendicular be erected both circles will 
touch it at the same point A. Therefore they touch each 
_ other at that point. In like manner it may be shown that 
NH, AFH, MF, AHF respectively touch at the points H, 
E. Therefore AFH is the circle required. 
By using OD—EM for OD+EM this demonstration is 
applicable to fig. 4, in which none of the circles are compre- 
hended by the given circle. 
2. When the touching circle comprehends avd touches 
one externally or comprehends one and touches two exter- 
nally. 
Cons. Let (Fig. 5. pl. 3.) HK, GM, LE, be the giv- 
en circles whose centres are A, B, C. Let EL be the cir- 
cle which is not to be circumscribed alone. From A with 
radius=radius of HK-+radius of LE describe the circle D 
I. And from B with radius of MG,+radius of LE, de- 
scribe the circle FN. Through C describe the circle DCF 
touching DI, FN in D, F,—of which circle let O be the 
centre. Decreasing the radius of this circle by a line=ra- 
dius of the circle LE (or in Fig. 6_ increasing it by the same 
line) describe the circle HEG, which shall be the circle re- 
quired. 
Demonstration. For join OC and it will cut the circle LE 
_ inthe pomt E. Because OR—OC—CE, E is in the cir- 
cumference of HEG. Therefore if from E a perpendicu- 
lar be erected, it will touch both circles in the same point 
E. Therefore they touch each other in that point. In like 
manner the circles G, M, HEG; HK, HEG, touch re- 
spectively at H, G. 
Now by using OC +CE for OC—CE this demonstration 
applicable to Fig. 6. pl. 3. in which one circle is compre- 
hended and the other two touched externally. 
Proziem XI. 
[t is required to draw a circle through a given point, to 
touch a straight line given in position and a circle given in 
magnitude and position. 
