If  tangent to the ellipse $x^2+4y^2=4$ meets the tangents at the extremities of its major axis at B and C, then the circle with BC as diameter passes through the point:

Equation of ellipse  $\frac{x^2}{4}+\frac{y^2}{1}=1$
Let contact point of tangent be  $(2\cos \theta ,\sin \theta )$
$\therefore$  Equation of tangent is  $\left(\cos \theta \right)x+2\sin \theta y=2-----\left(i\right)$
Put x = in (i) we get  $y=\frac{1-\cos \theta }{\sin \theta }=\tan \frac{\theta }{2}$
$\therefore C\equiv (2,\tan \frac{\theta }{2})$ 
Put x=-2 in (i), we get  $y=\frac{1+\cos \theta }{\sin \theta }=\cot \frac{\theta }{2}$
 $\therefore B\equiv (-2,\cot \frac{\theta }{2})$
Since, equation of circle whose ends of diameter are and  
 $(x_1,y_1)and(x_2,y_2)is(x-x_1)(x-x_2)+(y-y_1)(y-y_1)\equiv 0$
 $\Rightarrow (x+2)(x-2)+(y-\cot \frac{\theta }{2})(y-\tan \frac{\theta }{2})=0$

$x^2-4+y^2-(\tan \frac{\theta }{2}+co\mathrm{t}\frac{\theta }{2})y+1=0----\left(ii\right)$
So,  $\left(\sqrt{3,0}\right)$  satisfying option (ii)