<s>[INST] Find value of $x^2+y^2$ using given expression 
Consider two real numbers $x , y$ such that $$\left(x^2+1\right)\left(y^2+1\right)+9=6\left(x+y\right)$$ Hence find the value of $x^2+y^2$.

At first I tried to factorise the condition but the $(xy)^2$ created much problems. I also tried to create a complete square to get a simpler expression but to no avail. And just out of curiosity, can this question be interpreted geometrically so as to find the required value?
 [/INST] Hint. You can rewrite this as $$(y^2 + 1) x^2 - 6x + (y^2 - 6y + 10) = 0.$$
Taken as a quadratic function of $x$ this has discriminant $$36 - 4(y^2 + 1)(y^2 - 6y+10) = -4 (y^2 - 3y + 1)^2;$$ in particular, the only way to get a real solution is with $y^2 - 3y + 1 = 0$, i.e. $y = \frac{3 \pm \sqrt{5}}{2}$.
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