Official solution
\[{{{x^2}} \over 4} + {{{y^2}} \over 1} = 1\]
Equation of tangent i (cos\[\theta\])x + 2sin\[\theta\]y = 2
\[B\left( { - 2,{{1 + \cos \theta } \over {\sin \theta }}} \right),C\left( {2,{{1 - \cos \theta } \over {\sin \theta }}} \right)\]
\[B\left( { - 2,\cot {\theta \over 2}} \right)\]
\[C\left( {2,\tan {\theta \over 2}} \right)\]
Equation of circle is
\[(x + 2)(x - 2) + \left( {y - \cot {\theta \over 2}} \right)\left( {y - \tan {\theta \over 2}} \right) = 0\]
\[{x^2} - 4 + {y^2} - \left( {\tan {\theta \over 2} + \cot {\theta \over 2}} \right)y + 1 = 0\]
so, \[\left( {\sqrt 3 ,0} \right)\] satisfying option (1)