Strong's Problews. 274 
cles HGL, DIE meetin K. They also touch in that point. 
For if they do not they must meet in some other point ; let 
em meeting. Join Dx. Let Dx produced cut HGL in 
yand ABiny. Then Dy, De=LD, DK. But DaE be- 
ing aright angle as before, Dx, Dz =FD, DE=LD, DK. 
Therefore Dr, Dy=Dz, Dz and Dy=Dz which is absurd. 
Therefore the circles do not meet in z. Nowe any 
point they meet in no point but K. They sharaligee: touch 
in K. Wherfore LGKH is the circle required. 
_ Case IIL. When the given circle cuts the given straight 
Const. Let (Fig. 9. pl. 3.) AB, be the given straight 
line, H the given point, and FD’E ‘the given circle. Let 
the circle FD’E cut ABin L, M. Through C the oe 
of the given circle draw Cl at right sane io AB in 1. 
this line produced cut the circle FD’E in E, F. Through 
H, I, E deseribe the circle HIF. Join EH. Let this pro- 
duced cut G, HIF inG. Through H, G describe the circle 
GHOD touching AB in O. And this will be the circle re- 
juired. 
Se casioations For join OE, let EO extended cut E 
DF in D’ and GHOD in D. Then (E. 6. p. I. ) EO, E 
| O, ED. 
meet in D. But they also touch in this point. For if not, 
let them as before meet in z. Join Ex cutting AB in y, 
and GHOD. in z. Then joining Fa, the angle EzF isa 
right angle. Therefore El: Ez:: Ex: EF wherefore E 
I,EF=Ez, Ex. But EI, EF= EO, ED=EH, EG=ky, 
_Now z being any point but the point D, they meet in 
ae int but D. They therefore touch in D. Therefore 
GHOD is the circle required. 
Note—When the circle does not cut the line, the point 
must be given without the circle and on the same side 0 
the line with the circle. When the circle cuts the line the 
point may be given any where except at the point of inter- 
section of the line and circle. If the circle to be deseribed 
is to touch the given circle externally , the point may be giv- 
en any where without the circle or in the circumference, 
Taggers Se pee ts. E, F, Ae: al Case III.) scan are 
No. 
