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Compute the area of the circle that passes through all the intersection points of $4x^2 + 11y^2 = 29$ and $x^2 - 6y^2 = 6.$

 Dec 4, 2020
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Original graph  of the problem  : https://www.desmos.com/calculator/xyzfrdpqbv

 

Because of symmetry......we only need one intersection point

 

4x^2  +   11y^2    = 29      (1)

x^2    -    6y^2     =    6   →   -4x^2  + 24y^2  =  -24    (2)

 

Add (1)  and (2)   and we have

 

35y^2  =  5

y^2  = 5/35  =  1/7

y = sqrt (1/7)

 

And

 

x^2  - 6 (1/sqrt(7))^2  =  6

x^2   -  6/7   = 6

x^2  =  6 + 6/7

x^2  =  48/7

x  =  sqrt  (48/7)

 

(0,0)   will be the center of the  circle

 

The  radius =   sqrt (48/7  + 1/7) =   sqrt(  49/7)  =  sqrt (7)

 

The equation of the  circle we need =  x^2  + y^2  = 7

 

Area =   pi  (sqrt (7))^2  =   7 pi  units^2

 

Here's the completed graph   :   https://www.desmos.com/calculator/of0i8sikec

 

 

cool cool cool

 Dec 4, 2020

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