How to find the auxiliary circle of a non-standard ellipse?

Question

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.

Answer 1

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,

$$x^2+y^2=4$$

Answer 2

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.