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Two circles of radius r are externally tangent to each other and internally tangent to the ellipse x^2 + 2y^2 = 6 as shown below. Find r.

 

 Jun 2, 2021
 #1
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The  centers  of the circles will be  (r, 0)   and (-r, 0)

The  equations of the  circles are

(x -r)^2  + y^2  = r^2   ⇒  x^2  - 2xr  + r^2  +  y^2  =  r^2   ⇒  y^2  = 2xr - x^2

(x + r)^2 + y^2  = r^2  ⇒  x^2 + 2xr + r^2  +   y^2  = r^2     ⇒ y^2  = -2xr - x^2 =  -(2xr + x^2)

 

Subbing  in  y^2   in the equation of  the ellipse  we  have this system 

 

x^2  + 2(2xr  - x^2)  =  6

x^2  -  2 (2xr + x^2) =  6

 

x^2  + 4xr   -  2x^2  = 6

x^2  -  4xr   + 2x^2  = 6        add  these

 

2x^2  = 12

 

x^2  =  6

 

x = sqrt (6)

 

Using   x^2  + 4xr  - 2x^2 = 6  to solve for r

 

(sqrt (6))^2  + 4 (sqrt 6) r  - 2(sqrt 6)^2  =  6

6  +  4sqrt(6) r  -  12 =  6

4sqrt (6) r  = 12

sqrt (6)r   = 3

r =  3 / sqrt (6)   =  3/ ( sqrt 3  * sqrt 2)   =   sqrt  (3/2)

 

The  graph below shows  what we have.....the circles are   tangent to each other and  internally tangent to the ellipse   at the endpoints  of the major axis of the ellipse......

 

 

cool cool cool

 Jun 2, 2021

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