find value of a expression $\frac{(y_1-3)(y_2-3)}{x_1x_2}$ under some conditions

Question

If $x_1,y_1,x_2,y_2$ satisfy below equtions:$$x_1^2+y_1^2=9\quad (1)$$ $$x_2^2+y_2^2=9\quad (2)$$ $$(x_1+x_2)^2+(y_1+y_2-2)^2=4 \quad (3)$$ then find the value of the expression:$$\frac{(y_1-3)(y_2-3)}{x_1x_2}$$ the abolve equtions hava $4$ variables, so if we treat one variable as constant, for example:$x_1$. then solve other variables as the function of $x_1$: $y_1=f(x_1)$,$x_2=g(x_1)$,$y_2=h(x_1)$,even under the help of Mathematica. I can't do the job.because the expression if very large and have square root.
because $$\frac{(y_1-3)(y_2-3)}{x_1x_2}=\frac{y_1y_2-3(y_1+y_2)+9}{x_1x_2} \quad (4)$$ maybe find the relationship between $x_1x_2$,$y_1y_2$ and $y_1+y_2$ is the right way. If we add the above (1)(2)(3) equations, get $$9+x_1x_2+y_1y_2=2(y_1+y_2)$$ it can simplify the expression (4) a little.

Answer 1

(1)+(2): $x_1^2+x_2^2+y_1^2+y_2^2=18$

Putting in (2) we get:

$18+2(x_1x_2+y_1y_2)-6y_1-5y_2+9=4$

Rewriting as :

$2x_1x_2 +2(y_1y_2-3y_1-3y_2+9)=-5$

$2x_1x_2 +2(y_1-3)(y_2-3)=-5$

Dividing both sides by $x_1x_2$ we get:

$2+\frac{(y_1-3)(y_2-3)}{x_1x_2}=-\frac 5{x_1x_2}$

$A=\frac{(y_1-3)(y_2-3)}{x_1x_2}=-(2+\frac 5{x_1x_2})$

Now we can argue on value of A, for example if we want an integer:

$x_1x_2=5\Rightarrow A=-3$

A is a function of $x_1$ and $x_2$, so it can not have a fixed value.

Update: If :

(3): $(x_1+x_2)^2+(y_1+y_2-2)^2=4$

we have:

$18+2y_1y_2-2y_1-2y_2+2x_1x_2=0$

$y_1y_2-3y_1-3y_2+9=-x_1x_2-2(y_1+y_2)$

$(y_1-3)(y_2-3)=-x_1x_2-2(y_1+y_2)$

Dividing both sides by $x_1x_2$ we get:

$\frac{(y_1-3)(y_2-3)}{x_1x_2}=-(1+ 2\frac{y_1+y_2}{x_1x_2})$

Which is a function of $x_1, x_2, y_1$ and $y_2$.

Answer 2

I find a way, instead of evaluating $$\frac{(y_1-3)(y_2-3)}{x_1x_2}$$,i consider $$\frac{(y_1-3)^2(y_2-3)^2}{x_1^2x_2^2}$$ because it is easier to eliminate $x_1^2$,$x_2^2$, use equation $(1),(2)$,we have $$\frac{(y_1-3)^2(y_2-3)^2}{(9-y_1^2)(9-y_2^2)}=\frac{(y_1-3)(y_2-3)}{(y_1+3)(y_2+3)}$$ and use this $$9+x_1x_2+y_1y_2=2(y_1+y_2)$$ to get a relationship between $y_1+y_2$ and $y_1y_2$.rewriting it $$(x_1x_2)^2=[2(y_1+y_2)-y_1y_2-9]^2$$ wo we have : $$(9-y_1^2)(9-y_2^2)=[2(y_1+y_2)-y_1y_2-9]^2$$ so we have $$(y_1-3)(y_2-3)(y_1+3)(y_2+3)=[2(y_1+y_2)-y_1y_2-9]^2$$ $$\Rightarrow (y_1y_2-3(y_1+y_2)+9)(y_1y_2+3(y_1+y_2)+9)=[2(y_1+y_2)-y_1y_2-9]^2$$ in order to see clearly, we set $n=y_1y_2$,$m=y_1+y_2$,we only need to find the relationship between $m$ and $n$. after simplify, we have: $$m(-13m+4(n+9))=0$$,so we have $m=0$ or $13m=4(n+9)$, so the expression is equal to $-1$ or $\pm\frac{1}{5}$