Find value of $\lambda$ $S$ is a circle having center at $(0, a)$ and radius $b\lt a$. A variable circle centred at $(\alpha, 0)$ and touching the circle $S$ meets $X$ axis at $ M$ and $N$. A point $P=(0,\pm \lambda\sqrt {a^2-b^2})$ on the $y$ axis such that angle $MNP$ is a constant for any choice of $\alpha$ then find $\lambda$.
My work: 
I wrote down the equation of circle $S$ and I have tried using the cosine rule in the two triangles having the same constant angle $MNP$. I also tried using basic algebra and the Stewart's theorem in some of the triangles I could do work in.  But this was of no use. Moreover so many of variables are confusing me a lot.  Any help is greatly appreciated.
 A: There is no $\lambda$ such that $\angle MNP$ stays constant when $\alpha$ varies, since $\tan(\angle MNP)$ is a ratio of two segments, one of which is constant $\lambda \sqrt{a^2-b^2}\,$, while the other one clearly varies with $\alpha$.
The more interesting question, however, is whether such a $\lambda$ exists so that $\color{red}{\angle MPN}$ stays constant.
Let $c$ be the radius of the second circle, then the tangency condition gives:
$$\alpha^2 + a^2=(b+c)^2 \quad \iff \quad c = \sqrt{\alpha^2+a^2}-b$$
It follows that $M,N = (\alpha \pm c, 0)=\left(\alpha+b \pm \sqrt{\alpha^2+a^2}, 0\right)\,$, then with $p=\lambda\sqrt{a^2-b^2}\,$:
$$\require{cancel}
\begin{align}
\tan(\angle MPN) &= \tan(\angle OPN - \angle OPM) = \frac{\tan(\angle OPN)-\tan(\angle OPM)}{1+\tan(\angle OPN)\tan(\angle OPM)} \\[5px]
 &= \frac{\dfrac{\cancel{\alpha}+c}{p}-\dfrac{\cancel{\alpha}-c}{p}}{1+ \dfrac{\alpha+c}{p} \cdot \dfrac{\alpha-c}{p}} = \frac{2pc}{p^2+\alpha^2-c^2} \\[5px]
 &= \frac{2\lambda\sqrt{a^2-b^2}(\sqrt{\alpha^2+a^2}-b)}{\lambda^2(a^2-b^2)+ \cancel{\alpha^2}-(\cancel{\alpha^2}+a^2)+2b\sqrt{\alpha^2+a^2}-b^2} \\[5px]
 &= \frac{2\lambda\sqrt{a^2-b^2}(\sqrt{\alpha^2+a^2}-b)}{(\lambda^2-1)a^2-(\lambda^2+1)b^2+2b\sqrt{\alpha^2+a^2}} \\[5px]
 &= \frac{2\lambda\sqrt{a^2-b^2}}{2b} \,\cdot\, \frac{\sqrt{\alpha^2+a^2}\color{blue}{- b}}{\sqrt{\alpha^2+a^2}\color{blue}{-\dfrac{(1-\lambda^2)a^2+(1+\lambda^2)b^2}{2b}}}
\end{align}
$$
The latter does in fact not depend on $\alpha$ iff the $\color{blue}{\style{font-family:inherit}{\text{blue}}}$ terms are equal $\iff \lambda=\pm1\,$.
A: Folllwong @divx ‘s adjustment, the question is now “is whether such a $\lambda$ exists so that $\angle MPN$ stays constant.”
At first, we have circles $S_A$ and $S_T$ as given. [For simplicity, I took the liberty of letting the internal common tangent to pass through O. The logic will remain valid if there is an offset.]  
We get (1) Q = the point of contact; (2) AT = the line of centers; (3) OQ = part of L, the common tangent; and (4) $\angle OQT = 90^0$.

By power of a point, $OQ = \sqrt {OH.OK} = … = \sqrt {a^2 - b^2}$. We set $P_1$ be the first position of $P$ such that $OP_1 = OQ$. In other words, $P_1= (0, \sqrt{a^2- b^2})$ where $\lambda = +1$.
“$\alpha = 90^0 - \theta = 90^0 - \theta’ = \alpha_1 = \alpha_2$” implies $P_1QM$ is a straight line.
“$\alpha’ = \alpha_1 = \alpha$” implies $NQ \bot P_1QM$. This further means $ONQP_1$ is cyclic. Hence, $\angle NP_1M = \angle QON$, which is a constant.
For any other qualifying $P$ such that $\angle NPM = \angle NP_1M$, those P’s must lie on the segment $NP_1M$. At this stage, the y-axis will cut that circle at most two points and $P_1$ is already one. However, “$\alpha’’ = \alpha’ = \alpha_1$” implies the y-axis is tangent to that circle. This leaves $P_1 = P$ is the only candidate.
Symmetric argument shows there is a $P_1’$ when $\lambda = -1$.
