<s>[INST] Find the Minimum Value of $x^2+y^2$ Given:
$x+y=2\sin{a}-\cos{b}\\
xy=2\cos{a}+\sin{b}$
Find the minimum value of $x^2+y^2$.
Attempt:
$\begin{aligned}
x^2+y^2&=(x+y)^2-2xy\\
&=(2\sin{a}-\cos{b})^2-2(2\cos{a}+\sin{b})\\
&=4\sin^2{a}-4\sin{a}\cos{b}+\cos^2{b}-4\cos{a}-2\sin{b}
\end{aligned}$
$\begin{aligned}
f_{a}(a,b)&=0\\
8\sin{a}\cos{a}-4\cos{a}\cos{b}+4\sin{a}&=0\\
\cos{a}\cos{b}&=2\sin{a}\sin{b}+\sin{a}\\
\end{aligned}$
$\begin{aligned}
f_{b}(a,b)&=0\\
4\sin{a}\sin{b}-2\sin{b}\cos{b}-2\cos{b}&=0\\
2\sin{a}\sin{b}&=\sin{b}\cos{b}+\cos{b}\\
\end{aligned}$
Tried to plot it on a graph, the answer should be -6.

 [/INST] Write $b:={\pi\over2}+c$. Then the equations are
$$x+y=2\sin a+\sin c,\qquad xy=2\cos a+\cos c\ ,$$
and instead of $f$ you now have the function
$$g(a,c)=(2\sin a+\sin c)^2-4\cos a-2\cos c$$
representing $x^2+y^2$. One sees that for $a=c=0$ the function $g$ assumes the value $-6$, but it is impossible for $g$ to assume a smaller value. It follows that $f$ assumes the minimum $-6$ at $(a,b)=\bigl(0,{\pi\over2}\bigr)$.
Note that both $x$ and $y$ are complex at the extremal point. If it is an additional condition that $x$ and $y$ should be real we have a more difficult problem.
</s>