How to find the auxiliary circle of a non-standard ellipse? 
Given the equation of conic C is $5x^2 + 6xy + 5y^2 = 8$, find the equation S of its auxiliary circle?

Now, I know that the equation C is an ellipse.
Since $\Delta = 5*5(-8) - (-8)(3^2) \neq 0$
And,
$ab - h^2 = 5*5 - 3^2 > 0$
Which is the condition for an ellipse.
But this isn't a standard one!. In my school, we have only worked with ellipses of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. 
So, how do get the equation of the auxiliary circle for the given ellipse?
Any help would be appreciated.
 A: Since the conic is centre origin, we may use polar coordinates:
\begin{align}
  8 &= 5r^2\cos^2 \theta+6r^2\cos \theta \sin \theta+5r^2\sin^2 \theta \\
  r^2 &= \frac{8}{5+6\cos \theta \sin \theta} \\
  &= \frac{8}{5+3\sin 2\theta}
\end{align}
Now,
$$\frac{8}{5+3} \le r^2 \le \frac{8}{5-3} \implies 1 \le r^2 \le 4 \\$$

Hence, the auxiliary circle is $r=2$ or equivalently,
$$x^2+y^2=4$$

A: The standard way is that of switching to a pair of rotated axes: $X=(x-y)/\sqrt2$, $Y=(x+y)/\sqrt2$, but you probably haven't learned that yet.
As an alternate approach you can observe first of all that the center of the ellipse is $(0,0)$ and consider the intersections between the ellipse and a circle with the same center and radius $r$, with equation $x^2+y^2=r^2$. 
For a certain value of $r$ the circle touches the ellipse internally at two points, while for a greater value of $r$ it touches the ellipse externally at two points (and this one is precisely the auxiliary circle you must find). For all intermediate values of $r$, circle and ellipse intersect at four points. Hence you can find the auxiliary circle by searching for the values of $r$ which give only two intersections between the curves.
Subtracting the equations of circle and ellipse one gets
$$
6xy+5r^2=8,
\quad\text{that is:}\quad
y={8-5r^2\over6x}.
$$
Insert this into the circle equation, to eliminate $y$ and get an equation for $x$:
$$
36x^4-36r^2x^2+(8-5r^2)^2=0.
$$
This is a biquadratic equation, having two solutions when its discriminant vanishes:
$$
36^2r^4-4\cdot36\cdot(8-5r^2)^2=0.
$$
The larger solution of this equations gives then the radius of the auxiliary circle.
