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Strong’s Problems. 61 
the required circle is to circumscribe the given circle. See 
ig. 12. , 
Note. In case I. where the given lines are parallel, if 
ihe given circle and one of the given lines be on opposite 
sides of the other line, then the Problem becomes impos- 
sible. ; 
Prosuirm VII. 
To draw a straight line touching two circles given in mag- 
nitude and position. | 
ase I. When the touching line does not pass between 
their centres. 
Const. Let AF, BC (Fig. 13.) be the two given circles. 
Join their centres. Take CE = AB—BD, if AC> BD, 
and with CE radius and C (the centre of the given circle 
AF) centre, describe the circle EX. From D the centre of 
the other circle, draw DE touching the circle EX in E. 
Join CE, and produce CE until it meet the circle AF in 
the point A. At the point A draw the tangent AB and pro- 
duce it to the circle BG. ‘Then shall the line AB likewise 
touch the circle BG. 
Demonstration. For, ED being a tangent to the circle 
EX, the line CE drawn from the centre to the point of con- 
tact will be at right angles to ED.—For the same reason 
EC produced is at right angles to AB. Therefore ED, AB 
are parallel. BD, therefore, being drawn from the centre D 
perpendicular to AB ; ABDE will be a parallelogram, and 
EA, BD will be equal. But EA —radius of the circle 
BG. Therefore BD equaling radius of circle BG, the point 
_ B falls in the circumference of BG. And AB is at right 
angles to BH the radius of the circle BC in B. A 
therefore be a tangent to the circle BG in the point B. But 
AB is likewise a tangent to the circle AF in A, (by Const.) 
therefore AB is the tangent required. 
ase 11. When the touching line passes between the 
centres of the two given circles. 
Const. Let the two circles (Fig. 14.) be AB, DE.— 
From O the centre of the circle DE, draw OF = radius 
circle AB + radius circle DE, and with OF as radius describe 
the circle GF. From C the centre of the circle AB draw . 
CF touching FG in some point as F. Let the line joining 
