<s>[INST] Calculation of $\min$ distance between $x^2+y^2=9$ and $2x^2+10y^2+6xy=1$ 
Calculation of $\min$ distance between $x^2+y^2=9$ and $2x^2+10y^2+6xy=1$

$\bf{My\; Try::}$ Using the fact that the distance between two curve is independent of shifting.
So put $x=u+v$ and $y=u-v$ in thsese two curves, We get $\displaystyle u^2+v^2=\frac{9}{2}$
and $\displaystyle 2(u^2+v^2-2uv)+10(u^2+v^2-2uv)+6(u^2-v^2) = 1\Rightarrow 18u^2+6v^2-20uv=1$ 
But here again i get terms of $uv,$ So i did not understand how can i solve it, 
Help me, Thanks 
 [/INST] The original ellipse is congruent with the ellipse whose equation is $x^2+11y^2=1$.
And all the ellipse which is similar to it has the equation form:$x^2+11y^2=\lambda$ Make it and $x^2+y^2=9$ as a equation group and solve it: $ x = \sqrt \frac{99-\lambda}{10} or -\sqrt \frac{99-\lambda}{10}, y = \sqrt \frac{\lambda -9}{10} or - \sqrt \frac{\lambda -9}{10}$.That means when $ 9<=\lambda <= 99$, the ellipse has intersection points with circle.
Therefore, when $\lambda = 9$ the ellipse tangent with the circle.
The words are so abstract possibly. 
Therefore the minimal distance becomes distance between the long axis vertexes. Since one of the long vertexes of $x^2+11y^2=1$ is (-1,0) and one of the long vertexes of $x^2+11y^2=9$ is (-3,0). 
Hence, the answer should be 2.
Complement: 
1) why elipse: $x^2+11y^2=1$ is congruent with elipse:$2x^2+10y^2+6xy=1$.
Designate  $2x^2+10y^2+6xy=1$ as ellipse II.
Designate  $x^2+11y^2=1$ as ellipse I.
Think one of point$(x,y)$ on the ellipse II. Make it rotate about original point$(0,0)$ and it should fall into the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $. Find the rotation matrix:
$M = \left(\begin{matrix}cos(\theta)&-sin(\theta)\\sin(\theta)&cos(\theta)\end{matrix}\right)$
where $tan(\theta) = \frac{1}{3}$.
Using rotation matrix multiply the column matrix $ \left(\begin{matrix}x\\y\end{matrix}\right)$every point(x,y) on the ellipse II can get every point(u,v) on the ellipse I.
Now find the linear combination of $u^2, v^2$. Since $u^2 = f(x,y), v^2 = g(x,y)$ and the maximal degree of f(x,y) about x is 2. Similarly, the maximal degree of g(x,y) about x is 2. Through observing the coefficient, its equation should be $u^2 + 11 v ^2 = 1$.
2) Why choose $theta$ satisfying $tan(\theta) = \frac{1}{3}$?
Because our target is to make the ellipse formalized. Luckily, the ellipse's reflecting point is $(0,0)$ because there is no monomial item(i.e. x y) in the original equation. Therefore, assume a line: $y = kx$ passing through the ellipse II. Make $ y=kx$ and $ 2x^2+10y^2+6xy=1$ as a equation group. Solve it. Get $ x = \sqrt \frac{1}{10k^2+6k+2} y = \sqrt \frac{k^2}{10k^2+6k+2} $
Since the long vertex point of the ellipse has maximal distance on the ellipse about $(0,0)$, just find the $ \max x^2+y^2$. 


*

*$x^2+y^2 = \frac{k^2+1}{10k^2+6k+2}$ By the division law in differentiation, $(x^2+y^2)' = \frac{-18k-5}{(10k^2+6k+2)^2} = 0$. So when $ k = -\frac{1}{3} $, $x^2 + y^2$ has maximal.

*Since the rotation angle should be positive, and then $ tan(\theta) = \frac{1}{3}$


In a nutshell, the answer is 2.
If my answer has some errors or short points, please leave a comment. 
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