Tangent points of a circle touching two functions I'm really confused with this and don't quite know where to start, and was hoping that I could get a booster in the right direction!
See the picture, but here's a written description: A graph has two graphs plotted on it, $x=\ln(x)$ and $x=\ln(-x)$, to make a funnel shape. A circle is resting inside of that funnel, with a radius of $2$. How can I find a) the center point of the circle (which rests on the $y$-axis, I just need to know the $y$- value), and b) the tangent points or points of contact between the circle and the two plotted functions.
If you're looking at the graph, then it would be points $P$ and $Q$ for the tangent points and point $R$ for the center of the circle. 

Thank you!
BTW: please don't just give the answer! I just need a pointer on what to find and where to go from there.
 A: Start from the end and go back. 
Note that the segment $RQ$ is orthogonal to the tangent line to the graph of $\ln(x)$ at the point $Q$. So given a generic $x_0>0$, this the "action plan": 
1) Find the equation of tangent line at $Q=(x_0,\ln(x_0))$. 
2) Find the line orthogonal at $Q$ to the tangent line. 
3) Determine the intersection point $R$ of the orthogonal line and the line $x=0$.
4) Now that we have the $y$-value of $R$ as a function of $x_0$, find $x_0>0$ such that the the distance between $R$ and $Q$ is $2$.
Can you take it from here?
A: HINT:
I think the equations should be
$$y=ln(x)........(1)$$
$$y=ln(-x)........(2)$$
$$(x-h)^2+(y-k)^2=4........(3)$$
Now at tangent point the slope of the circle and the given graph will be the same.so,
$$\dfrac{d}{dx}\bigg((x-h)^2+(y-k)^2-4\bigg)=\dfrac{d}{dx}\bigg(ln(\pm x)\bigg)........(4)$$
You will get a system of equations here..4 equations,4 unknown...
Just solve them.
A: HINTS:
Draw normal/tangent at $Q$ cutting symmetry line at $R$
Coordinates $$ Q:\quad (x_1, \log \,x_1) $$
Slope is the derivative evaluated at that point.
$$ \tan \phi= 1/x =1/x_1 $$
Difference along central symmetry axis, vertical component of $QR$
$$ h = x_1 \cot \phi $$
Coordinates $$ R: (0,\, \log x_1 +h) $$
Equation of circle tangent at $ (P,Q) $ centered at $R:$
$$ (x- 0)^2 + [y-(\log\, x_1 +h)]^2 = (h/\sin \phi)^2 $$
