Question

If a 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

• A. $(-1,1)$
• B. $(\sqrt{3},0)$
• C. $(1,1)$
• D. $(\sqrt{2},0)$

Answer 1

The correct option is B $(\sqrt{3},0)$

Tangent at $(2\cos \theta ,\sin \theta )$ is
$\frac{x\left(\cos \theta \right)}{2}+\frac{y\left(\sin \theta \right)}{1}=1$
Coordinates of $B(-2,\cot \frac{\theta }{2})$ and $C(2,\tan \frac{\theta }{2})$
Now, the equation of circle with $BC$ as diameter is
$(x-2)(x+2)+(y-\cot \frac{\theta }{2})(y-\tan \frac{\theta }{2})$
$\Rightarrow x^2+y^2-y(\tan \frac{\theta }{2}+\cot \frac{\theta }{2})-3=0$
At $y=0,x=\pm \sqrt{3}$
Hence, circle passes through the point $(\sqrt{3},0)$.