60 - Strong’s Problems. 
respectively.—Jom ON. Let ON cut AB in L. Then with 
O as centre and OL as radius, describe a circle KLP ; 
which shall be the circle required. i 
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 isa 
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 OQ and ON being radii of the circle 
QNS are equal. Suppose the lme QO to meet the given 
circle in X. Then (by Const.) QX == NL. Therefore 
OL =OX. Hence the circle LP passes through X. And 
if at the point X a perpendicular were erected, it would be 
a tangent to both cireles at the same point X. The cireles 
therefore touch each other at the point X. Wherefore 
XP is the circle required. 
Case Il. When the two given straight lines mtersect 
‘each other, and the cirele is given in any position. 
Construction. Let AB, CD (Fig. 11.) be the given straight 
lines and SN the given circle. It is required to describe a 
circle to touch AB, CD, and the given circle. Draw EO, 
OG, 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. JomMF. Let MF 
eut AB in X. Then from M as centre with radius MX, 
describe a circle. And this shall be the circle required.— 
Join MN intersecting the circle SV in 8. 
Demonst. For NM, MF being radii 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 KO, and EO being 
parallel to AB, it is likewise perpendicular to AB. There- 
fore AB isa tangent to the circle SXW.. In hke manner we 
may prove that CD touches SXW. Now, if from the point 
S a perpendicular be drawn to NM, it will be a tangent to 
both circles at the same point. Therefore the circles SXW, 
SV touch each other in S, whence SXW is the circle re- 
uired. 
; By using NM-+NS for MN—SN and MX+XF for 
MX—XI", the above demonstration is applicable where 
