268 Strong’s Problems. 
Case IV. When the two circles are equal, and the 
touching circle circumscribes both or neither of them. 
Const. Let (Fig. 4. pl. 2.) CKL, BMN be the given 
circles, and A the given point. From A as centre with ra- 
dius—radius of the given circles, describe IQ. Then 
through az, y, the centres of the given circles describe the 
circle I+ y touching IQ in the point 1. Let O be the cen- 
tre of this circle. From O as centre, and (in Fig. 1.) O 
I+1IA as radius describe the circle CAB, which shall be the 
circle required. 
Dem. For join OyB=OI+I1A. Therefore the circle 
ABC weets the circle MN, in the point B.. But it likewise 
touches this circle in the point B. For at B draw FG at 
right angles to OyB. Because this line is at right angles to 
the diameter of ABC in the point B, it touches this circle in 
the pomt B. For the same reason, it touches MN in B. 
Therefore the circle ABC touches the circle MN in the 
point B. In like manner it may be shown, that ABC 
touches KLC in the point C, and it passes (by Const.) 
through A. ABC is therefore the eirele required. 
Now (Fig. 5. pl. 2.) by using OI—1A for OJ+IA, the 
construction and demonstration employed in Fig. 4, are ap- 
plicable to Fig. 5, in which the given circles are neither of 
them cir cumscribed by the touching circle. 
Cor. 1. In Case I, when the given circles touch, that 
is, when C, D ede the circle LCD must be described 
touching AE in the point of coincidence of C, D, after 
which, the construction and proof are as before. 
Cor. 2. In Cases I, I, if the two circles cut each other, 
the solution remains the same ; for it is independent of the 
distance of their centres. But in Case IL]. when the cir- 
cles cut each other, the problem becomes impossible. 
When they touch, the circle touching them must pass 
through the point of contact of the two circles. 
Cor. 3. Case IV, may be considered as falling under 
Cases I, II, when the point A (See Fig. Cases J, IT.) be- 
comes infinitely distant. But in Case III, the construction 
remains the same, whatever be the magnitude of the circles. 
For there the point A is confined between the centres of 
the circles. 
Cor. 4. In Case I, if the given point fall in the line AE 
between the points C, D, as in r, make the rectangle Ar. 
