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ABCD is a square with side length 1.  A circle passes through B, C, and is tangent to the midpoint of AD (at E).  What is the radius of the circle?

 

 

 Dec 21, 2020
 #1
avatar+114325 
+1

Here's one way to  solve this....but....maybe not the fastest....!!!!

 

Let D  =(0,0)    E  =  (0, 1/2)    B  =(1.1)  C  = (1,0)

 

Call  the center of the circle  ( m, 1/2)

 

So  the distances from each of these points to the center  are equal  = the radius

 

We have this 

 

Distance from center to E   = distance from center to B

 

( m - 0)^2  + ( 1/2 - 1/2)^2  =  ( m - 1)^2 + ( 1/2 -1)^2

 

m^2  =  m^2 - 2m + 1  + 1/4

2m = 5/4

m = 5/8

 

So  the radius  =   the distance from (m,n)  to E  =  

 

sqrt  [ 0 - 5/8)^2  +  ( 1/2 - 1/2)^2 ] =

 

sqrt[  ( 5/8 )^2 ] =  

 

5/8

 

 

cool cool cool

 Dec 21, 2020
 #2
avatar+859 
+1

ABCD is a square with a side length of 1.  A circle passes through B, C, and is tangent to the midpoint of AD (at E).  What is the radius of the circle?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

AB = EF = 1      BF = 1/2         ∠BEF = tan-1(1/2) = 26.56505118º

 

EB = sqrt(EF2 + BF2) = √1.25

 

Radius      EO = 1/2 (EB / cos ∠BEF) = 0.625     or   5/8

 Dec 22, 2020

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