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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.

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,
$$dfrac{d}{dx}bigg((x-h)^2+(y-k)^2-4bigg)=dfrac{d}{dx}bigg(ln(pm x)bigg)........(4)$$

You will get a system of equations here..4 equations,4 unknown...
Just solve them.

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 $R:$

$$ (x- 0)^2 + [y-(log, x_1 +h)]^2 = (h/sin phi)^2 $$

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 $x_0>0$, find the tangent line at $Q=(x_0,ln(x_0))$, the orthogonal line and determine $R$ which is along the line $x=0$.
The final step is to find $x_0>0$ such that $|RQ|=2$.

Can you take it from here?