Given two intersecting circles finding the coordinates of intersection of common tangent? 
Q) A circle $C_{1}$ is drawn having point P on x-axis as its centre
  and passing through the centre of the circle $C:x^2 +y^2=1$. A common
  tangent to $C_{1}$ and $C$ touches the circles at Q and R
  respectively . Then $Q(x,y)$ always satisfies $x^{2}=\lambda $ , then
  find $\lambda$ ?

Attempt
Let $(p,0) $ be the centre of $C_1$ then we have $C_1 = x^2 +y^2 -2px=0$. 
Let $R=(x_1 ,y_1 )$ and $Q=(x_2 , y_2 )$ 
Then I wrote the equation of tangents of both circles and equated them and got 
$\frac{1}{p}=x_1 + x_2$ 
How do I proceed? Hints?
 A: $\boldsymbol{r\lt1}$
Using similar triangles, we get the $x$-coordinate of $Q$ to be $r+(1-r)=1$.

The sum of the orange and lavender segments times $r$ should be the length of the lavender segment; that is,
$$
\frac{\color{#C00000}{r}}{\color{#00A000}{1}}(\color{#FF8000}{r}+\color{#8080FF}{x})=\color{#8080FF}{x}
$$
solving gives
$$
x=\frac{r^2}{1-r}
$$

$\boldsymbol{r\gt1}$
Using similar triangles, we get the $x$-coordinate of $Q$ to be $r-(r-1)=1$.

The sum of the lavender and orange segments should be $r$ times the length of the lavender segment; that is,
$$
\color{#8080FF}{x}+\color{#FF8000}{r}=\frac{\color{#C00000}{r}}{\color{#00A000}{1}}\color{#8080FF}{x}
$$
solving gives
$$
x=\frac{r}{r-1}
$$

If the center of $C_1$ is on the left of $(0,0)$, then the $x$-coordinate of $Q$ will be $-1$.
A: Slope form of tangent equation of circle with centre $(a,b)$, slope $m$ and radius $r$ is: $(y-b)=m(x-h)+r\sqrt{1+m^2}$
let tangent to $C$ be: $y=mx+\sqrt{1+m^2}$
Tangent to $C_1$: $y=m(x-h)+h\sqrt{1+m^2}$
$(h,0)$ is the centre of $C_1$
Since the tangents represent the same line, $$-mh+h\sqrt{1+m^2}=\sqrt{1+m^2}$$
Solving the above equation for m, $$m=\pm\frac{h-1}{\sqrt{2h-1}}$$
I will take the plus symbol for $m$. ($\pm$ shows two such tangents are possible.)
Substituting $m$ in the the tangent equation of $C$ and rearranging, $$(y\sqrt{2h-1}+x)-h(x+1)=0$$
This is an equation of family of straight lines passing through point of intersection of lines $y\sqrt{2h-1}+x=0$ and $x=-1$ with $h$ as the parameter. It's clear that $Q$ has to be the point of intersection of the family of straight lines since that's the only point that lies on the the line for any value of $h$.
The x-coordinate of point of intersection is $-1$. So $x^2=1$.
