Strong’s Problems. 5% 
Through D draw DE at right angles to AE. Produce DE 
until EF = ED. Through the points D, F’, describe a cir- 
cle to touch the line AB.* And this shall be the circle re- 
quired. : 
Demons. For suppose the circle EGH to touch the le 
AB in the point G, '‘Yhrough G draw Gh at right angles to 
AB, and cutting the Ime 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 I'D at right angles it 
likewise passes through the centre. 1, must therefore be the 
eentre. From L. draw LH perpendicular to AK. Now, 
since angle LAG — angle LAH, and the angle AGL = an- 
gle LHA, and Al is common to both triangles, LG = LH, 
The circle, therefore, passes through the point H. And since 
LH is at right angles io AC, the circle FGH touches the 
line AC. But (by Construction) it touches AB, and passes 
through the point D; FGH is therefore the circle required. 
Q@ EL. 
Prospiem V. 
Tt is required through two given points to describe a cir- 
cle which shall touch a circle, given in position and magni- 
tude. 
Case 1. 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.—It is required to describe a circle 
such, that it shall pass through the points B, C (B, C’) and 
touch the given circle. Jom BC. Bisect BC nn D. Take 
F, the centre of the circle AB. Join BF. Through D, draw 
DE at right angles to CB, meeting BF produced 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. Itis also manifest 
* Problem I. 
Vou. Il.....No. 1. g 
