<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] Let $2=r\cos C,1=r\sin C, r\ge0\implies r^2=5$
$$x^2+y^2$$
$$=(2\sin A-\cos B)^2-2(2\cos A+\sin B)$$
$$=r^2\sin^2(A-C)-2r\cos(A-C)$$
$$=r^2+1-\left(r\cos(A-C)+1\right)^2$$
Now for $r\ge0,$
$$(r+1)^2\ge\left(r\cos(A-C)+1\right)^2\ge0$$
</s>