378 Strong’s Problems. 
respectively equi-distant from the points of intérsection of 
the given circle and line. 
Proguirm XU. 
It is required to describe a circle to touch a straight line 
given in position, and two circles given in position and 
magnitude. 
Case I. When the touching circle circumseribes both 
the given circle and touches the straight line, or circum- 
scribes neither of the given circles and touches the given 
straight line. 
Const. Let (Fig. 10, pl. 3.) AB be the given straight 
line, GQS, MPO, the given circles, it is required to de- 
scribe a circle to touch at B and likewise touch the circles 
~GQS, MPO. Let I and N be the centre of the given cir- 
cles. From [, with radius=radius of RGS= rable of M 
PO (if G@S > MPO) deseribe the circle HRK. Draw 
also the line CD parallel to AB, and distant from i by a 
line=radius of the circle MPO. ‘Then through N describe 
the circle NHF touching HRK in H and CD io F.. Let 1, 
be the centre of this circle. From L with radius=radius 
of the circle HE N-tradius of the circle MPO describe the 
circle EGO, which will be the circle required. 
— Demonstration. For jom LAG which will pass through 
I. Now because LG= LH + radius of the circle MPO and 
IG=HI-+ radius of the circle MPO, G is in the circumfer- 
ences of EGO, SGQ. And if at G a perpendicular be 
erected, it will touch both circles EGO, 5G@Q at the point 
G. Therefore these circles touch each other at the point 
G. In like manner it may be shown that the circles EGO, 
MPO touch at the point O. But EGO likewise touches 
the straight line AB. For join LFE. Let this line cut the 
circle EGO in E. New because LE=LE + radius of the 
circle MPO, the point E falis in the line AB. And be- 
eause AB and CD are parallel (LEA=/LFC. But 4L 
’ FC isa right angle—(F being the point of contact of the 
eircle FHN and line ED) Therefore LEA is a right an- 
gle, and consequently AES touches EGO in E, wherefore 
EGO is the circle required. 
By using LE +radius of MPO for LE—radius of MPO, 
this demonstration is applicable to Fig. 11, when neither of 
the circles is comprehended. 
