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A circle entirely in the first quadrant is tangent to the y-axis and to the line y= x/2, and the center of the circle lies on the line y = 2.  Find the radius of the circle.

 May 16, 2020
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Since the center of  the circle lies  on the  line y = 2  it must have  the  coordinates  ( a, 2)

 

The point of tangency wth the y axis  = (0,2)

 

The line can be written   as  x - 2y  =  0

 

Using the formula for the distance from a point to a line  and equating this with the distance from(a, 2) to (0,2) = a  we can solve for  "a" thusly

 

l  (a) - 2(2) l

___________  =     a

sqrt [ 1 + 2^2]

 

   l a - 4 l

_______   =   a

  sqrt (5)

 

l a - 4 l =    a sqrt (5) 

 

Then either

 

a - 4 = asqrt (5)     

 

a - asqrt (5)  = 4

 

a ( 1 - sqrt (5))  = 4

 

a =  4 /( 1 -sqrt (5) )      reject this...."a"  is negative

 

 

Or

 

4 - a  =  a sqrt (5)

 

4  =  a + asqrt (5)

 

4  = a ( 1 + sqrt (5))

 

a  =  4  / (1 + sqrt  (5) )   =   the radius

 

Here's a graph :  https://www.desmos.com/calculator/gijqact2cn   

 

 

cool cool cool

 May 16, 2020

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