Solve a system of equation 
Solve this system of equations.
  $$\begin{cases}
x^{2} + y^{2} = 4 \\ 
z^{2} + t^{2} = 9 \\
xt + yz = 6
\end{cases}$$

My attempts: 


*

*$(x^{2}+y^{2})(z^{2} + t^{2})=36 \xrightarrow{\text{by applying Lagrange formula}} (xt + zy)^{2} + (zx -ty)^{2}=36 \xrightarrow{xt + yz = 6}  36 + (zx -ty)^{2} =36\Rightarrow (zx -ty)^{2} = 0 \Rightarrow zx = ty \Rightarrow z = \frac{ty}{x}$

*$xt + yz = 6 \xrightarrow{z = \frac{ty}{x}}  xt + \frac{ty^{2}}{x} = 6 \Rightarrow x^{2}t + y^{2}t = 6x \Rightarrow t(x^{2} + y^{2})=6x \xrightarrow{x^{2} + y^{2} = 4} t=\frac{3}{2}x$

*$z= \frac{ty}{x} \xrightarrow{t=\frac{3}{2}x} z = \frac{3}{2}y$ 
Can we go further and determine the exact value of variables?
 A: We have $4$ unknowns with three equations. So, we should not have deterministic solutions
$zx=ty\implies \dfrac xy=\dfrac tz=k$(say)
$$4=y^2(1+k^2)$$ $$9=z^2(k^2+1)$$
$$\text{and }6=yz(k^2+1)$$
$$yz\ne0,\dfrac96=\dfrac{z^2}{yz}=\dfrac zy$$
$\implies \dfrac z3=\dfrac y2=u$(say)
$$\implies\{x,y,z,t\}=\{k(2u),2u, 3u,k(3u)\}$$  where $u(\ne0),k$ are arbitrary numbers such that $$4=4u^2(1+k^2)$$
$$u^2(k^2+1)=1$$
A: $$\left(\frac{x}{2}\right)^2+\left(\frac{y}{2}\right)^2=1$$
$$\left(\frac{t}{3}\right)^2+\left(\frac{z}{3}\right)^2=1$$
Let $x_1 = \frac{x}2, y_1=\frac{y}2, t_1=\frac{t}{3}, z_1 = \frac{z}3$.
Then $P=(x_1, y_1)$ and $Q=(t_1, z_1)$ are points on the unit circle.
Also, from the third constraint,  we know that $\cos \angle POQ= 1 $
Hence $\frac{x}2=\frac{t}{3}$ and $\frac{y}{2}=\frac{z}{3}$ and they need to satisfy $\left(\frac{x}{2}\right)^2+\left(\frac{y}{2}\right)^2=1$. The answer is not unique. 
We can charactherize it by a parameter $\theta$ where $\cos(\theta)=\frac{x}2=\frac{t}{3}$ and $\sin(\theta)=\frac{y}{2}=\frac{z}{3}$
A: You have $4$ variable and $3$ equation. So one of the variables is free and you can assign some value to it, for example $y=1$, so you have $z=\frac{3}{2}$ , $x=\sqrt{3}$ , $t=\frac{3}{2}\sqrt{3}$.  
