376 Strong’s Problems. 
Case 1. When the given circle is not comprehended. 
Const. Let (Mg. 7 pl. 3.) AB be the given straight 
line, H the given point and DIKE the given circle. It is 
required to describe through H, a circle to touch the given 
line and circle. Let C be the centre of DIKE. From C 
draw CF at right angles to AB, cutting the circumference 
of DIKE in E, D. Through F, E, H describe the circle 
FEH (Prob. I.) jon DH. Suppose DH produced to cut 
the circle FEH in G. Through GH describe the circle 
HGL to touch AB in L (Prob. LHI.) and this will be the 
circle required. 
Demonstration. For join OL. Let OL cut the circle 
LGH in K and DIK’E in K’. Now DE, DF=DH, DG 
=DK, DL. Hf EK’ be joined, (Plaf. Euc. 6. prop. L,) B 
E, DF—DL, DK’; but BE, DF—DL, DK therefore DL. 
DK—DL, DK’. Hence KD=DK’. Therefore the cir- 
cles DIKE LKG meet in E. But they likewise touch in 
that point. Forif they do not they must meet.in some oth- 
er point. Let them meet ina. Join Dx and extend it to 
cut L.HG in y, and AB in z. Then as before, (Kuclid. 6. 
p. 1,) Dx, Dz =DE, DF =DL, DK=Dz, Dy. Therefore 
since De, Dz=Dza, Dy, Dz=Dy, the less to the greater 
which is absurd. Therefore the circles do not meet in any 
point but K. Wherefore they touch in that point. But 
(by Const.) GKL touches AB and passes through H. G 
KL is therefore the circle required. 
Case If. When the given circle is circumscribed. 
Const. Let (Fig. 8. pl. 3.) AB be the given line, DIEK 
the given circle and H the given point. From C the cen- 
tre of DIEK draw CF perpendicular to AB, cutting the 
given circle in D, E of which E is not adjacent to the 
straight line. Through F, E, H describe the circle FEH. 
Jom HD. Let HD extended cut FEH in G. Thr ough 
G, H describe a circle touching AB in the point L. and 
this will be the circle required. 
Demonstration. For join LD extended let this line cut 
LGH in the point K and EDI in the point K’.. Now FD, 
DE=LD, DK. But if KE be joined the angle DKE is a 
right angle, and the angles KDE and FDL are equal. 
Therefore FD: LD::K’D: : DE, whence FD, DE=L 
D, DK’ which therefore “is equal to LD, DK. Hence D 
K'=DK and the points K’, K, coincide. Therefore the cir- 
