Strong’s Problems. 57 
‘Through D draw DE at right angles to AE. Produce DE 
until EF = ED. Through the points D, F, deseribe a cir- 
cle a touch the line AB.* And this shall ba the circle re- 
quin 
Fsinupe. For suppose the circle EGH to touch the line 
AB in the point G. Through G draw GL at right angles to 
AB, and cutting the line AE in L. Because GL is drawn 
at right angles to the tangent AB, it passes through the cen- 
tre, and since AE bisects the chord FD at right angles it 
likewise passes through the centre. L must therefore be the 
centre. From L draw LH perpendicular to AE. Now, 
since angle LAG = angle LAH, and the angle AGL = an- 
gle garde and AL is common to both triangles, 1 LG= ee 
Prosuem V. 
It is required through two given points to describe a cir- 
cle which shall touch a circle, given in position and magni- 
tude. 
Case I. When one of the given points is in the circum- 
ference of the given circle and the other either within or 
without the given circle. 
-, Const. Let AB (Fig. 7.) be the given circle, B the point 
in the circumference, and C, (or C-) the point without (or 
within) the given circle. wot is required to describe a circle 
such, that it shall pass through the points B, C (B, C*) and 
touch the given circle. Join BC. Bise ct BC in D. Take 
F’, the centre of the circle AB. Join BF. "Through D, draw 
DE at right angles to CB, meeting BF produced i in E. Join 
CE, and with E as centre, and radius BE, describe the circle 
CB; ; then will CB be the circle required. 
Demonstration. Because CD = DB, and the angle CDE 
=angle BDE, and DE is common to both the triangles CDE, 
BDE, CE=BE. Therefore the circle described from E as 
centre, with radius BE passes through C. It is also manifest 
* Problem H. 
Wh Rb cecist Wo: 2. g 
