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Segment AB=6 is tangent to the circle with center O. Lengths of other segments are BC=8, CD=10, and OD=12.  Find the radius of the circle.

 

 Jan 17, 2021
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 Jan 17, 2021
 #2
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Let E  be  the intersection  of  BC with the  circle

 

By  the secant tangent theorem,   

 

BE (  BE + EC)  = AB^2

 

BE  ( 8) =  6^2

 

BE (8)  =  36

 

BE  = 36/8   =  18/4  =  4.5

 

So   chord CE  = 8  - 4.5  =  3.5

 

If  we call the midpoint of this chord, F  and we  draw   OF....then triangle    ODF is right with OD the hypotenuse

And  DF  =  10 + CF =  10  + CE/2  =  10 + 3.5/2   =   11.75

 

And by  the  Pythagorean Theorem,   OF^2  = OD^2  -  DF^2  = 5.9375

 

And we can  draw another  right triangle  OFC  with  the radius  of the  circle, r, as a hypotenuse, legs CF  and  OF

 

So

 

r^2  =  CF^2 + OF^2

 

r^2  =  (1.75)^2  + 5.9375

 

r^2   =  9

 

r =  3

 

cool cool cool

 Jan 17, 2021

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