62 Strong’s Problems. 
O, F cut the ae DEinD. From D draw DA parallel 
to FC. From C draw CA parallel to OF and let it cut 
DA Sidaced. in A. Then will DA be the tangent re- 
uired. 
= Demonstration. For because CF touches the circle FG 
and from O the centre of FG, is drawn to the point of 
contact, the angle OFC is a right angle. But DA is paral- 
lel to FC and is therefore perpendicular to OF. Hence it 
touches DE. And AC being parallel to DF is at right an- 
les to 
; Moreover the ap ae ACFD isa parallelogram, and there- 
fore AC= DF. But DF =radius of the circle AB.— 
Therefore A is in i Gronabiesey Oo . Now, the an- 
gle DAC has been proved a right angle. Wherefore DA 
‘ouches the circle AB in the point A. But it likewise 
pb oe ee circle ED. AD is therefore the tangent re- 
uire % = 
ig Or: t to Case I. When the circles become ets that is, 
parallel to DC the line ne joining the centre re of a tre circles. 
Cor. to Case Il. When the circles becom e equal, that 
is, when OD = AC, OF — 20D, therefore == 20X, 
X being in the middle of the line OC. 
Note. That this problem is impossible i in both Cases, 
when one circle lies wholly within the other; in scar’ il, 
when one circle cuts ‘the other. 
Prosiem VIII. 
it is required to find a point, from which any straight 
lines being drawn, cutting two circles given in magnitude 
and position shall cut off similar segments. 
sef. When the point does not fall between the two 
circles 
Const.* Let BD and PE (Fig. 15. ) wi two circles. 
Draw BPA touching the circles in B an Prob. VII. 
Case I.) and produce this tangent, to meet TG (which } = 
ihe centres of the given circles) in some pointas A. From 
Both the cases of this Problem admit of avery sipigle construction, 
which is independent ofthe 7th. A line joining the extremities of any two 
radii phot parallel to each other, will is gba ie! line joining the centres 
(produced, in Case I.) in the point required — 
