60 Strong’s Problems. 
Se cel oin ON. Let ON cut ABi in L. Then with 
O as centre and OL as radius, describe a circle XLP; 
which shall ie the circle required. . 
Demons. For ON by the nature of the tangent is_per- 
pendicular to EF, and therefore to AB, which is parallel to 
EF. Now since XPL passes through L, and ALO is a 
right angle, XPL must touch AB in the point L. In like 
manner it may be proved to touch CD in P. But it like- 
wise touches the given circle. For, join QO the centres of 
the two circles. Then O@ and ON being radii of the circle 
QNS are Sani 5 OT “ line QO to eee the given 
circle in Then (by Const.) QX =NL. Therefore 
OL= Ox. Hence the eels LP passes ee X. And 
if'at the point X a perpendicular were erected, it would be 
a tangent to both circles at the same point 2 he circles 
therefore touch each other at the point Xx. Wherefore 
XLP is the circle required. 
Case Il. When the two given straight mee intersect 
each other, and the circle is given in any posi 
Construction. Let AB, cD (Fig. 11.) be the tae en straight 
- lines and SN the given circle. It is required to describe a. 
circle to touch AB, CD, and the given circle. Draw EO, 
%, parallel to the two given lines and respectively distant 
from them by a line = radius of the given circle-—Let N 
be the centre of the given circle. ‘Through N describe a 
circle NZ touching the lines EO, OG in the points F, F ; 
of which circle let M be the centre. Join MF. Let MF 
cut AB in X. Then from M as centre with radius MX, 
describe a circle. And this shall be the circle required.— 
in 8. 
oin MN intersecting the circle § 
Demonst. For NM, MF roe sai of the same circle 
are equal. But NS = XF (by Const.) therefore SM = 
MX. Therefore the circle MW passes through the point 
S. Now MXF being perpendicular to EO, and EO being 
parallel to AB, it is likewise perpendicular to ‘AB. There= 
fore AB isa tangent to ic 8 circle SXW. In like manner we 
may prove that CD touches $ Now, if from the point 
S a perpendicular be pom to NM, it will be a tangent to 
both circles at the same point. Therefore the circles SKW, 
SV wi cach other in S, whence SXW is the cirele re- 
uire 
‘ By ‘using NM-+WNS for MN—SN and MX+XF for 
MX—XF, the above demonstration is applicable where 
i, —_- lll ta eile atndiiem 
