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AB is a chord of the circle x^2 + y^2 = r^2 and the tangents at A and B meet at C. If M(1,2) is the midpoint of AB and the area of quadrialteral OACB is sqrt(11/5)*r^2, find the radius r of the circle.

 

 Dec 14, 2020
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OM = sqrt  (1^2  + 2^2)  = sqrt (5)

 

Because OCB is a right triangle  with CBO  =  90 °

 

Then BM  is a geometric mean  such that

 

   CM/BM  = BM / OM

 

BM^2  = CMsqrt (5)

 

CM =  BM^2 /sqrt (5) 

 

And   OACB  forms a  kite

 

Its area  =(1/2) product of the  diagonal lengths

 

So

 

(1/2)  ( BM + BM )  ( OM + CM)

 

(1/2) ( 2BM)  (  sqrt (5) + BM^2 / sqrt (5) )

 

BM sqrt (5)  +  BM^3 / sqrt (5)  =   sqrt (11/5) r^2

 

r ^2  =  [ BM sqrt (5)  + BM^3 /sqrt (5) ] / sqrt (11/5)

 

r^2  =  OM^2  + BM^2

 

[ BM sqrt (5)  + BM^3 /sqrt (5) ] / sqrt (11/5)  =  5 + BM^2

 

sqrt (5)  [ BM sqrt (5)  + BM^3 / sqrt (5) ]  / sqrt (11)  = 5 + BM^2

 

5 (BM) + BM^3   = 5sqrt (11)  + BM^2 sqrt (11)

 

5(BM)  + BM^2 (BM)  = 5 (sqrt (11)) + BM^2 (sqrt (11) )

 

Note  that  this will be true when   BM  = sqrt (11)

 

r = sqrt  ( BM^2  + OM^2) =  sqrt  ( 5 + 11)   =sqrt (16)  = 4

 

 

cool cool cool

 Dec 14, 2020
edited by CPhill  Dec 14, 2020
edited by CPhill  Dec 15, 2020

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