Australian Math Competition Geometry Problem 

$PQRS$ is a rectangle with a centre $C$. $PQ$ has length $4$ and $PS$ has length 12. The circles meet $PS$ at $U$ and $V$ with both having radius $1$. $PU$ has length $1$ and $PV$ has length $4$. What is $PW$?

I’ve tried this problem for days and tried to find answers elsewhere but I can’t do it. Tried getting areas of triangles to find altitudes and was working with trapeziums and such. Made lots of approaches but I always get stuck.

Draw a rectangle where C is extended straight downwards until it reaches PS (call this point of intersection A) and one line from C to the left until it reaches PQ. This rectangle has side lengths 2 and 6, because if C is the center of rectangle PQRS, then it should divide the rectangle's sides into halves as well. Also, notice that it is tangent to both the circles because they have the same height as this new rectangle.
From there, we can calculate that the distance from the rightmost point of the right circle to line CA is 1 because 6-1 (the radius of the circle)=1. Extend lines AP and BC one unit to the left, and create a larger rectangle. Notice now that if you extend line CW one unit to the left as well, it intersects perfectly with the left bottom corner of the rectangle. Call this point W'. 
As per the definition of a diagonal, this newly extended line CW' divides the new rectangle exactly in half. And given that these circles are positioned so that they are horizontally symmetrical, it is clear why the original line CW divided the areas into halves. 
Now for the final calculation, we take the dimensions of this new larger rectangle. Its height is 2 and its width is 7. Therefore, the slope of the diagonal Cw' is 2/7. Taking point W' as the origin of a coordinate system, the coordinates of point W is (1, 2/7) and the coordinates of point P is (1, 0). It should then be clear that the distance is 2/7!
Sorry if the formatting or logic seems a bit janky; I'm a high schooler in South Korea and this is new for me :/
