266 Strong’s Problems. 
MATHEMATICS. 
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Ant. XI. Mathematical Problems, with Geometrical Con- 
structions and Demonstrations, by Professor 'THroporr 
STRONG. 
[Continued from page 64 of this Volume.} 
Propiem IX. 
Ir is required through a given point to describe a circle 
which shall touch two circles given in position and magni- 
tude. 
Case I. When the two circles are unequal, and the cir- 
cle which touches them does not circumscribe them. 
Const. Let L (Fig. 1. pl. 2.) be the given point, and 
HBe, HG, E the given circles. It is required to describe 
through L, a circle which shall touch the two given circles. 
Join the centres x, y, of the circles HBC, DGE by cy, and 
extend xy till it meets FG, (FG being drawn, (Prob. vu. 
Case i.) touching the two circles,) in A. Let vy extended, 
cut the given circles in B, C, D, E. Through L the given 
point, and C, D, the two adjacent points, in which AK cuts 
the given circles, describe (Prob. i.) the circle LCD. Join 
LA, and suppose LA produced cuts LCD in Kk. Through 
the points L, K, describe (by Problem v.) a circle touching 
HBC in H. And this shall be the circle required. 
Demonstrations. For join AH, and extend it till it meets 
the circle DG in I. It will meet this circle, because it cuts 
off similar segments from the two given circles, (Prob. viii.) 
And let AH meet the circle HLK in I’. Now by the prop- 
erty of the circle AL, KA=AC, AD. But AC, AD=A 
H, Al, (Prob. viii. Cor. 4.). Therefore AH, AI—=AL, 
LK. But AL, AK—=AH, Al’. Therefore AH, AIl/=AH, 
AJ. Hence (striking out AH) AI==AI’. Wherefore the 
points I’, I coincide. Therefore the circle LKH, meets 
. the circle DG, E in [. 
It also touches it in this point; for if the line MO be 
drawn touching KLH, BRH in H, and the line No. 2 touch- 
ing the circle LHI in I, then the angle RHM-—=angle in the 
