62 : Strong’s Problems. 
O, F cut the circle DE in D. From D draw DA parallel 
toFC. From C draw CA parallel to OF and let it cut 
DA produced in A. Then will DA be the tangent re- 
quired. 
Demonstration. For because CF touches the circle FG 
and from O the centre of FG, OF is drawn to the point of 
contact, the angle OFC is aright 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- 
gles to DA. : 
Moreover the figure ACF'D is a parallelogram, and there- 
fore AC=DF. But DF =radius of the circle AB.— 
Therefore A is in the circumference of AB. Now, the an- 
gle DAC has been proved a right angle. Wherefore DA 
touches the circle AB in the point A. But it likewise 
touches the circle ED. AD is therefore the tangent re- 
quired.— Q. E. I. 
Cor. to Case I. When the circles become equal, that is, 
when BD = AE, EC disappears. And BA is manifestly 
parallel to DC the line joining the centre of the two circles. 
Cor. to Case I]. When the circles become equal, that 
is, when OD = AC, OF — 20D, therefore OC = 20X, 
X being in the middle of the line OC. 
Note. That this problem is impossible in both Cases, 
when one circle lies wholly within the other; in Case IJ, 
when one circle cuts the other. 
Prosuem VIII. 
It is required to find a pomt, from which any straight 
lines being drawn, cutting two circles given in magnitude 
and position shall cut off similar segments. 
Case I. When the point does not fail between the twe 
circles. 
Const.* Let BD and PE (fig. 15.) be the two circles. 
Draw BPA touching the circles in B and P (Prob. VII. 
Case I.) and produce this tangent, to meet FG (which joins 
the centres of the given circles) in some pointas A. From 
* Both the cases of this Problem admit ef avery simple construction, 
which is independent ofthe 7th. A line joining the extremities of any twe 
radii drawn parallel to each other, will intersect the line Joiming the centres 
(produced, in Case I.) in the point required. —Edit. 
