276 Strong’s Problems. 
Case I. When the given circle is not comprehended. 
Const. Let (Fig. 7 pl. 3.) AB be the given straight 
line, H the given point ‘and DIKE the given circle. Iti is 
line and circle. Let C be “the centre of DIKE. From C 
— CF at right angles to AB, cutting the circumference 
in E, D. Through F, E, H “describe the circle 
F EH (Prod. £.) join DH. Suppose DH produced to cut 
the circle FEH in G. Through GH describe the circle | 
HGL to touch AB inL (¢ Prob. IIL. ) and this will be the 
circle required. 
emonstration. For joi n OL. Let OL cut the circle 
LGH in K and DIKE in Ki. Now DE, DF=DH, DG 
=DK, DL. If EK’ be joined, (Plaf. Eue. 6. prop. L,) B 
E, DF=DL, DK’; but BE, DF==DL, DK therefore DL. 
DE-=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. t them meet inz. Join Dx and extend it to 
cut LHG in y, and AB in z. Then as before, (Euclid. 6. 
p. 1,) Dr, Dz =DE, DF=DL, DK=Dz, Dy. Th erefore 
D 
She is absurd. ‘Therefore the circles do not meet in any 
int but K. Wherefore they touch in that point. But 
{by Cons. GKL touches AB and passes eign H. G 
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 
t line. Through F, E, H describe the circle FEH. 
Join deo Let HD extended cut whoop in G.- a 
right angle, and the angles E and FDL are eq 
ue FD: : LD: :K’D: $7: DE; —_ FD, DE= 
