Suppose $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$. What are $x,y,z,t$?

Question

Let $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$.

What are $x,y,z,t$?

Answer 1

Consider $$ (x^2+y^2)(z^2+t^2)-(xt+yz)^2=0 $$ Expanding the left-hand side you get $$ x^2z^2+\color{red}{y^2z^2}+\color{red}{x^2t^2}+y^2t^2 -\color{red}{x^2t^2}-2xyzt-\color{red}{y^2z^2}=0 $$ that simplifies to $$ (xz-yt)^2=0 $$ so $xz=yt$.

Now consider the linear system in the unknowns $z$ and $t$: \begin{cases} yz+xt=6 \\ xz-yt=0 \end{cases} With standard methods you get $$ z=\frac{6y}{x^2+y^2}=\frac{3}{2}y, \qquad t=\frac{6x}{x^2+y^2}=\frac{3}{2}x $$ Now any pair of numbers $x,y$ with $x^2+y^2=4$ will do.

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Alternatively, consider the complex numbers $$ A=x+yi,\qquad B=z+ti $$ Then the hypotheses yield $|A|=2$, $|B|=3$ and that the imaginary part of $AB$ is $6$, that is $$ AB=C+6i $$ with $C$ real. Then $\bar{A}\bar{B}=C-6i$ and multiplying gives $$ A\bar{A}B\bar{B}=C^2+36 $$ Thus $36=C^2+36$ and so $C=0$. Therefore $$ AB=6i $$ and $$ z+ti=\frac{6i}{x+yi}=\frac{6i}{x-yi}{4}=\frac{3}{2}i(x-yi)=\frac{3}{2}(y+xi) $$ whence again $z=\frac{3}{2}y$ and $t=\frac{3}{2}x$.

Answer 2

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 $(1)$.

Answer 3

Putting the parametric equation for both circles $$x=2\sin a,y=2\cos a,z=3\sin b,t=3\cos b\\$$From this you get$$xt-yz=6\sin a\cos b+6\sin b\cos a=6\sin(a+b)=6\\\sin(a+b)=1\\a+b=\frac{\pi}{2}+2k \pi$$ Which implies $x=2\sin(\pi/2-b)=2\cos b$ and $y=2\cos(\pi/2-b)=2\sin b$ From this we see that $(x,y,z,t)=(2\cos b,2\sin b,3\sin b,3\cos b)$ for each $b$