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.

We begin with the ellipse

$$x^{2}+4y^{2}=4.$$

It can be rewritten as $$\dfrac{x^{2}}{4}+\dfrac{y^{2}}{1}=1,$$ so we have

$$a^{2}=4,\qquad b^{2}=1,\qquad a=2,\qquad b=1.$$

Because $$a\gt b,$$ the major axis is the $$x$$-axis and its extremities are the end points

$$A_{1}(2,0)\quad\text{and}\quad A_{2}(-2,0).$$

For an ellipse $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1,$$ the tangent at a point $$(x_{1},y_{1})$$ is

$$\dfrac{xx_{1}}{a^{2}}+\dfrac{yy_{1}}{b^{2}}=1.$$

At $$(2,0)$$ this gives $$\dfrac{x\cdot2}{4}=1\;\Longrightarrow\;x=2,$$ and at $$(-2,0)$$ it gives $$\dfrac{x(-2)}{4}=1\;\Longrightarrow\;x=-2.$$

Thus the tangents at the extremities of the major axis are the two vertical lines

$$x=2\qquad\text{and}\qquad x=-2.$$

Now let us draw an arbitrary tangent to the ellipse that has slope $$m$$. For the ellipse in slope form the tangent is

$$y=mx\pm\sqrt{a^{2}m^{2}+b^{2}},$$

and with $$a^{2}=4,\;b^{2}=1$$ this becomes

$$y=mx\pm\sqrt{4m^{2}+1}.$$

Denote the quantity $$D=\sqrt{4m^{2}+1},\qquad D\gt 0.$$

Choosing the ‘$$+$$’ sign (the ‘$$-$$’ sign gives the same straight line shifted in $$y$$ by an overall minus and therefore the same pair of points $$B,C$$), the tangent is

$$y=mx+D.$$

We next find its intersections with the two fixed vertical lines.

Intersection with $$x=2$$ :

Substitute $$x=2$$ in $$y=mx+D$$ to obtain

$$y_{B}=m\cdot2+D=2m+D.$$

Hence $$B(2,\;2m+D).$$

Intersection with $$x=-2$$ :

Substitute $$x=-2$$ to get

$$y_{C}=m(-2)+D=-2m+D.$$

Hence $$C(-2,\;-2m+D).$$

Thus the endpoints of segment $$BC$$ are

$$B(2,\;2m+D),\qquad C(-2,\;-2m+D).$$

We must study the circle having $$BC$$ as its diameter. A very useful fact is the following:

For a segment with endpoints $$(x_{1},y_{1})$$ and $$(x_{2},y_{2})$$, a point $$(x,y)$$ lies on the circle with that segment as diameter if and only if

$$ (x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})=0. $$

This is simply the algebraic translation of “$$\angle BPC$$ is a right angle” (the dot‐product of the vectors $$\overrightarrow{PB}$$ and $$\overrightarrow{PC}$$ must be zero).

Applying the criterion with $$B(2,2m+D)$$ and $$C(-2,-2m+D)$$, we get that a point $$P(x,y)$$ will be on the required circle iff

$$$ \bigl(x-2\bigr)\bigl(x+2\bigr)+\bigl(y-(2m+D)\bigr)\bigl(y-(-2m+D)\bigr)=0. $$$

The first bracket simplifies immediately:

$$ (x-2)(x+2)=x^{2}-4. $$

For the second product we write the two subtractions explicitly and use the identity $$(A-B)(A+B)=A^{2}-B^{2}$$:

$$$ \begin{aligned} (y-2m-D)(y+2m-D) &=(y-D-2m)(y-D+2m)\\ &=(y-D)^{2}-(2m)^{2}\\ &=(y-D)^{2}-4m^{2}. \end{aligned} $$$

Putting both parts together, the condition for $$P(x,y)$$ becomes

$$ x^{2}-4+\bigl(y-D\bigr)^{2}-4m^{2}=0. \quad -(★)$$

We now check each of the four option points one by one. Remember that $$D=\sqrt{4m^{2}+1}.$$

Option A $$\bigl(\sqrt{3},0\bigr)$$ :

Insert $$x=\sqrt{3},\;y=0$$ into (★).

First term: $$x^{2}-4=(\sqrt{3})^{2}-4=3-4=-1.$$

Second term: $$\bigl(y-D\bigr)^{2}=(0-D)^{2}=D^{2}=4m^{2}+1.$$

Hence the left side of (★) is

$$$ -1+(4m^{2}+1)-4m^{2}=0. $$$

It is identically zero for every real value of the slope $$m$$. Therefore the point $$(\sqrt{3},0)$$ always lies on the circle whose diameter is $$BC$$.

Option B $$\bigl(\sqrt{2},0\bigr)$$ :

Put $$x=\sqrt{2},\;y=0$$ in (★).

$$x^{2}-4=2-4=-2$$.

Second term remains $$D^{2}=4m^{2}+1$$.

Total $$-2+(4m^{2}+1)-4m^{2}=-1\neq0,$$ so $$(\sqrt{2},0)$$ is not on the required circle.

Option C $$(1,1)$$ :

$$x^{2}-4=1-4=-3.$$

$$(y-D)^{2}=(1-D)^{2}=1-2D+D^{2}=1-2D+4m^{2}+1=2-2D+4m^{2}.$$

Total $$-3+\bigl(2-2D+4m^{2}\bigr)-4m^{2}=-1-2D\neq0.$$

So $$(1,1)$$ is not on the circle.

Option D $$(-1,1)$$ :

$$x^{2}-4=1-4=-3.$$

$$(y-D)^{2}=(1-D)^{2}$$ is the same as above, giving $$2-2D+4m^{2}.$$

Total $$-3+\bigl(2-2D+4m^{2}\bigr)-4m^{2}=-1-2D\neq0.$$

Therefore $$(-1,1)$$ is also excluded.

Only the first option satisfies the circle condition for every tangent, so the circle with $$BC$$ as diameter always passes through the point $$(\sqrt{3},0)$$.

Hence, the correct answer is Option A.