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.

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 $(0,-\frac{5}{3})$.
