How prove this $BC$ always passes through a fixed point with $\frac{x^2}{4}+y^2=1$ 
if the point $A(0,1)$ on the ellipse $\Gamma:$ $\dfrac{x^2}{4}+y^2=1$ and  the circle $\tau:$ $(x+1)^2+y^2=r^2(0<r<1)$,if $AB,AC$ tangent the circle $\tau$ ,$B,C\in \Gamma$,show that the line $BC$ always passes through a fixed point


I try Let $AB:y=kx+1$ then 
$$\begin{cases}
\dfrac{x^2}{4}+y^2=1\\
y=kx+1
\end{cases}
$$
so we have
$$x^2+4(kx+1)^2=4\Longrightarrow (4k^2+1)x^2+8kx=0$$
so we have
$$B(-\dfrac{8k}{4k^2+1},\dfrac{1-4k^2}{1+4k^2})$$
But for $C$ it hard to find it.
 A: The equation of the line passing through $A(0,1)$ is given by $mx-y+1=0$. 
Since we want this line to be tangent to the circle, we have
$$r=\frac{|m(-1)-0+1|}{\sqrt{m^2+(-1)^2}},$$
i.e.
$$(r^2-1)m^2+2m+r^2-1=0\implies m_1+m_2=\frac{2}{1-r^2},\quad m_1m_2=1\tag1$$
Eliminating $y$ from $mx-y+1=0$ and $x^2/4+y^2=1$ gives
$$\frac{x^2}{4}+(mx+1)^2=1\implies x(x+4m^2x+8m)=0\implies x=0,\frac{-8m}{4m^2+1}$$
and
$$y=mx+1=m\cdot \frac{-8m}{4m^2+1}+1=\frac{-4m^2+1}{4m^2+1}$$
So we can write $B\left(\frac{-8m_1}{4m_1^2+1},\frac{-4m_1^2+1}{4m_1^2+1}\right),C\left(\frac{-8m_2}{4m_2^2+1},\frac{-4m_2^2+1}{4m_2^2+1}\right)$. 
Using $(1)$, we have
$$\frac{b_y-c_y}{b_x-c_x}=\frac{\frac{-4m_1^2+1}{4m_1^2+1}-\frac{-4m_2^2+1}{4m_2^2+1}}{\frac{-8m_1}{4m_1^2+1}-\frac{-8m_2}{4m_2^2+1}}=-\frac{m_1+m_2}{3}=\frac{2}{3(r^2-1)}$$and
$$\frac{-c_xb_y+b_xc_y}{b_x-c_x}=\frac{-\frac{-8m_2}{4m_2^2+1}\cdot\frac{-4m_1^2+1}{4m_1^2+1}+\frac{-8m_1}{4m_1^2+1}\cdot\frac{-4m_2^2+1}{4m_2^2+1}}{\frac{-8m_1}{4m_1^2+1}-\frac{-8m_2}{4m_2^2+1}}=-\frac{5}{3}$$
So, the equation of the line $BC$ is given by
$$y=\frac{b_y-c_y}{b_x-c_x}x+\frac{-c_xb_y+b_xc_y}{b_x-c_x}=\frac{2}{3(r^2-1)}x-\frac 53$$
which passes through $(0,-5/3)$.
A: When $r\rightarrow 1$, it's easy to see that $y$-axis will be a tangent line, which means that the fixed point should be on the $y$-axis. So later, we need to put $x=0$ in the equation of tangent line to find the fixed $y$-coordinate. Use the tangent conditions, get $\frac{|k-1|}{\sqrt{k^2+1}}=r$ (here, we get $k_1$ and $k_2$), that is, $(1-r^2)k^2-2k+1-r^2=0$, or
\begin{eqnarray*}
k^2=\frac{r^2+2k-1}{1-r^2}.
\end{eqnarray*} 
Use Vita's law ($k_1k_2=1$ and $k_1+k_2=\frac{2}{1-r^2}$) to simply the equations of $AB$, you can get
\begin{eqnarray*}
y=-\frac{2}{3(1-r^2)}(x+\frac{8k_1}{4k_1^2+1})-1+\frac{2}{4k_1^2+1}.
\end{eqnarray*}
Set $x=0$, use the first equation to substitute $k^2$ in the above equation and simply, you will get a constant $-\frac{2}{3}-1=-\frac{5}{3}$. So $AB$ must pass through the point $(0,-\frac{5}{3})$.
