<s>[INST] Suppose $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$. What are $x,y,z,t$? Let $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$.
What are $x,y,z,t$?
 [/INST] Presumably, this should be solved over $\Bbb R$.
In this case the easiest way to think about this is that you have two (real) vectors $v = (x,y)$ and $w = (t,z)$.
Then,
$$\|v\| = \sqrt{x^2+y^2} = \sqrt 4 = 2,$$
$$\|w\| = \sqrt{t^2+z^2} = \sqrt 9 = 3,$$
$$v\cdot w = xt + yz = 6.$$
If we denote with $\varphi$ the angle between $v$ and $w$, we have
$$\cos\varphi =\frac{v\cdot w}{\|v\|\|w\|} = 1\implies \varphi = 0\implies w =\lambda v,\ \lambda>0.$$
Since $\|w\| = \lambda\| v\|$, we conclude that $\lambda = \frac 32$.
Considering polar coordinates, we can write $v = r(\cos\alpha,\sin\alpha)$, $w = \frac 32 r(\cos\alpha,\sin\alpha)$, where $\alpha\in[0,2\pi)$. Note that $$\|v\| =r\sqrt{\cos^2\alpha + \sin^2\alpha} = r\implies r = 2,$$ meaning that $v = 2(\cos\alpha,\sin\alpha)$, $w = 3(\cos\alpha,\sin\alpha)$, or
\begin{array}{cc}
\begin{align}
x &= 2\cos\alpha,\\
y &= 2\sin\alpha,\\
t &= 3\cos\alpha,\\
z &= 3\sin\alpha.
\end{align}&\tag{1}\end{array}
Returning to the original system we get
$$x^2+y^2 = 4\cos^2\alpha+4\sin^2\alpha = 4,$$
$$t^2+z^2 = 9\cos^2\alpha+9\sin^2\alpha = 9,$$
$$xt+yz = 6\cos^2\alpha + 6\sin^2\alpha = 6,$$
and thus, all (real) solutions are given by $(1)$.
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