Strong’s Problems. 59 
C, D, describe the circle ACD, which shall be the circle 
required. Through A draw FG perpendicular to EH. 
Demons. Because CE= ED, HC = HD and HE is 
common to the triangles HED, HEC, the angles CEH, 
DEH are equal, being opposite equal sides. ‘Therefore 
HE is perpendicular to DC. Now because CD isa chord in 
the circle CAD and is bisected at right anges by AE, AE 
passes through the centre of the circle. But FG is at right 
angles to KA, and EA passes through the centre of the cir- 
cle CDA ; therefore FG touches the circle CDA in the 
point A. "But (by Const.) FG touches the circle AB in the 
point A. Therefore the circles CDA, AB, touch each other 
at the point A.— 
In like manner a using the letters C-, D-, &c. for C, D, 
&c. the above demonstration is applicable to the case where 
se aaa are within. the circle at equal distances from the 
pSbhoHuc As CD, GF are both at right angles to EH 
they are parallel to each other. Therefore the construction 
in Case II, failing, Case ILI is necessar 
Note I. When one of the points is within the circle and 
the other without, the problem becomes impossible ; for 
then the circle which passes through those points will cut 
the given circle, which is against the Hy 
ote II. All the cases of this problem (except the first) 
admit of two solutions ; as is manifest from the above con- 
struction. 
Prosiem VI. 
It is required to describe a circle to touch two given 
aieaignt lines and a given circ 
Case I. When the two eye straight lines are parallel 
and the given circle lies between them, or cuts one or both 
of them 
Const. Let AB, CD (Fig. 10.) be the two given straight 
lines, and MI the given circle. Draw EF parallel to AB 
and distant from it, by a line = radius of the given circle. 
Draw also GH Sarallek to. CD and ata like distance from CD. 
It is _: to be noted that if EF fall between the given lines 
GH must likewise. Through Q the centre of the given 
circle cee the pores QNS touching EF, GH in N, S, 
