If $|z+2|-|z-2|=2$ and $\dfrac{z-a}{z+a}=ik$, where $k$ is a real parameter, $a\in \mathbb R$ has exactly two solution. Then interval of $a$ is? 

If $|z+2|-|z-2|=2$ and $\dfrac{z-a}{z+a}=ik$, where $k$ is a real parameter, $a\in \mathbb R$ has exactly two solution. Then interval of $a$ is?


My Approach:
First Curve is one branch of a hyperbola with its transverse axis of length $2$ and conjugate axis with length of $2\sqrt3$.
So hyperbola is $\dfrac{x^2}{1}-\dfrac{y^2}{3}=1$.
And second curve is circle $(k^2-1)x^2+(k^2-1)y^2+2a(k^2+1)x+(k^2-1)a^2=0$ with center $\bigg(\dfrac{-a(k^2+1)}{k^2-1},0\bigg)$  and radius $\bigg|\dfrac{2ak}{k^2-1}\bigg|$.
Then I put $y^2=3(x^2-1)$ obtained from hyperbola into circle $(k^2-1)x^2+(k^2-1)y^2+2a(k^2+1)x+(k^2-1)a^2=0$
Then I obtained the equation $4(k^2-1)x^2+2a(k^2+1)x+(k^2-1)(a^2-3)$
Now I don't Know how to proceed further?
Second Doubt
What if there was two branches of hyperbola for example $\bigg ||z+2|-|z-2|\bigg |=2$?
 A: Well, it is not hard to show that:
$$\frac{\text{z}-\text{a}}{\text{z}+\text{a}}=i\text{k}\space\Longrightarrow\space\begin{cases}
\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)-\text{a}^2=0\\
\\
\text{k}=\frac{2\text{a}\Im\left(\text{z}\right)}{\Im^2\left(\text{z}\right)+\left(\Re\left(\text{z}\right)+\text{a}\right)^2}
\end{cases}\tag1$$
We also see that:
$$\left|\text{z}+2\right|-\left|\text{z}-2\right|=2\space\Longrightarrow\space\sqrt{\left(\Re\left(\text{z}\right)+2\right)^2+\Im^2\left(\text{z}\right)}-\sqrt{\left(\Re\left(\text{z}\right)-2\right)^2+\Im^2\left(\text{z}\right)}=2\tag2$$
Now, we can solve the following equation:
$$\sqrt{\left(\Re\left(\text{z}\right)+2\right)^2+\text{a}^2-\Re^2\left(\text{z}\right)}-\sqrt{\left(\Re\left(\text{z}\right)-2\right)^2+\text{a}^2-\Re^2\left(\text{z}\right)}=2\space\Longrightarrow\space$$
$$\Re\left(\text{z}\right)=\frac{\sqrt{3+\text{a}^2}}{2}\tag3$$
And:
$$\frac{3+\text{a}^2}{4}+\Im^2\left(\text{z}\right)-\text{a}^2=0\space\Longleftrightarrow\space\Im\left(\text{z}\right)=\pm\frac{\sqrt{3\left(\text{a}^2-1\right)}}{2}\tag4$$
And we can also see that:
$$\text{k}=\frac{2\text{a}\left(\pm\frac{\sqrt{3\left(\text{a}^2-1\right)}}{2}\right)}{\left(\pm\frac{\sqrt{3\left(\text{a}^2-1\right)}}{2}\right)^2+\left(\frac{\sqrt{3+\text{a}^2}}{2}+\text{a}\right)^2}=\pm\frac{\sqrt{3\left(\text{a}^2-1\right)}}{\sqrt{3+\text{a}^2}+2\text{a}}\tag5$$
