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A circle lies inside a quarter-circle, as shown below.  The circle is tangent to side $\overline{AO}$ and arc $AB$ and side $\overline{BO}$.  If the radius of the quarter-circle si 2, then find the radius of the circle.

 

 Jan 7, 2024
 #1
avatar+128448 
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Let X  be the point of tangency  to the edges of  both circles  

 

Let the other two tangent points at the intersection points of the radius of the quarter-circle and the edge of the small circle be Y and Z

 

Let the center of the small circle  = M

 

Connect  MY , MZ and MO

 

Triangles MYO  and MZO  form congruent  45 - 45 - 90 right triangles

 

Let the radius of the small circle  = r

 

MZ =  r     MO =  (sqrt2) * r

 

Connect OX

 

OX  = MO  +  MX

 

OX =  2       MO = sqrt2 * r      MX  = r

 

So

 

MO + MX  = OX

 

sqrt (2)r + r =  2

 

r ( sqrt 2  + 1)  =  2

 

r =  2 / (sqrt (2)   + 1)  ≈  .828

 

 

cool cool cool

 Jan 7, 2024
edited by CPhill  Jan 7, 2024

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