Minimum Value of $(u-v)^2$ and $\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$

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anemone
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Here is this week's POTW:

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Find the minimum value of $(u-v)^2+\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$ for $0<u<\sqrt{2}$ and $v>0$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
on Phys.org
No one answered last week's POTW.(Sadface)

Below is a suggested solution:
The given function is the square of the distance between a point of the quarter of circle $x^2+y^2=2$ in the open first quadrant and a point of the half hyperbola $xy=9$ in that quadrant. The tangents to the curves at (1, 1) and (3, 3) separate the curves, and both are perpendicular to $x=y$, so those points are at the minimum distance, hence the answer is $(3-1)^2+(3-1)^2=8$.
 

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