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 [Q. Then 
through a, y, the centres of the given circles describe the 
cirele [+y touching IQ in the point I. Let O be the cen- 
tre of this circle. From O as centre, and (in Fig. 1.) O 
I+1JA as radius describe the circle CAB, which shall be the 
circle required. 
Dem. For join OyB=OI+IA. Therefore the circle 
ABC meets the circle MN, in the point B. But it likewise 
touches this circle in the point B. For-at B draw 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 point 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 
touches KLC in the point C, and it passes (by Const.) 
through A. ABC is therefore the circle required. 
Now (Fig. 5. pl. 2.) by using OL—IA for O1+IA, the 
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 I, U1.) 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 0! 
the circles. 3 
or. 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. 
