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.
+128474 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
