$x^2+y^2=1$, find min/max of $(3x+2y)^2+(x+2y)^2$ I have been trying to solve this problem,

For $x, y \in \mathbb{R}$ such that $x^2+y^2=1$, find the minimum and maximum value of $$(3x+2y)^2+(x+2y)^2$$

There are many ways to solve this problem such as 


*

*Trigonometry (substitute $x = \sin\theta, y = \cos\theta, \theta\in\left[0, 2\pi\right)$)

*Lagrange Multipliers

*Solve for $y$ and plug it in to the wanted expression and differentiate it.

*If we let the wanted expression equal to some constant $k$, it is an ellipse thus use linear transformation to rotate and find values of $k$ such that the ellipse is circumscribed inside/outside the circle
Using trigonometry, I found out that the answer is min: $9-\sqrt{65}$, max: $9+\sqrt{65}$.
But this problem actually came up in an algebra textbook (no relation to calculus or even pre-calculus), with the topic of Quadratic Equations and Discriminants.
Considering the context of the textbook, I am trying to find a solution using discriminants.
My first try was to let the given expression $(3x+2y)^2+(x+2y)^2 = k$ and assume that the min/max will happen when the discriminant is greater than or equal to $0$. (Since the equation must have real roots).
So what I did was expand to get $10x^2+16xy+8y^2=k$, use $x^2+y^2=1$ to simplify into
$$2x^2+16xy+8-k=0$$
this equation of $x$ must have real roots so $D/4\geq0$.
$$64y^2-2(8-k)\geq0$$
Now I am currently stuck here. Is there a way to get around this? I would really like to find a solution that uses discriminants. Thanks in advance.

Note: I just mentioned the $4$ other ways to solve this problem so that I can emphasize that I would really like to see a solution using discriminants. 
Also, any solution without calculus (or with only topics covered until high school - maybe -) are always welcome.
 A: Let
$$f(x,y)=(3x+2y)^2+(x+2y)^2=10x^2+16xy+8y^2.$$
If $f(x,y)=a$ for some $x$, $y$ on the unit circle then
$$f(x,y)-a(x^2+y^2)=0\tag1$$
has nonzero real solutions. But if $(1)$ has a nonzero real solution,
one can scale it so that $x^2+y^2=1$, and then $f(x,y)=a$ for that $(x,y)$.
But
$$f(x,y)-a(x^2+y^2)
=(10-a)x^2+16xy+(8-a)y^2.$$
This has a nonzero real solution iff the discriminant
$$16^2-4(10-a)(8-a)\ge0.$$
Solve this inequality for $a$ to get the values of $f$ on the unit
circle.
A: $$y=10\sin^2t+16\sin t\cos t+8\cos^2t$$
Divide both sides by $\cos^2t$ and set $\tan t=a$
$$(1+a^2)y=10a^2+16a+8\iff a^2(10-y)+16a+8-y=0$$ which is a Quadratic Equation in $y$
So, the discriminant must be $\ge0$
$$\implies16^2\ge4(10-y)(8-y)\iff(y-9)^2\le64+1\iff-\sqrt{65}\le y-9\le\sqrt{65}$$
A: Substituting with trig functions gives,
$$y=10\sin^2t + 16\sin t\cos t+8\cos^2t$$
Using the identity $\sin^2x+\cos^2x=1$, it simplifies to
$$y=8 + 2\sin^2t + 16\sin t \cos t$$
Use the double-angle formula $\sin{2x}=2\sin x \cos x$ and half-angle formula $\sin^2t = \frac{1-\cos{2t}}{2}$.
$$y=9+8\sin 2t-\cos 2t$$
Now,
$$y = 9+\sqrt{65}\left(\frac{8}{\sqrt{65}} \sin 2t - \frac{1}{\sqrt{65}} \cos 2t\right)$$
$$y=9+\sqrt{65} \sin{(2t-\alpha)}$$
where $\cos\alpha=8/\sqrt{65}, \sin\alpha=1/\sqrt{65}$
Thus maximum is $9+\sqrt{65}$, minimum is $9-\sqrt{65}$.
