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?

Continuing from Dylan's answer, $xy = 1$ and $x+y = 3$. Then:
$$x^2 + y^2 = (x+y)^2 - 2xy = (3)^2 - 2(1) = \boxed{7}.$$
Moreover, note that we can swap the values of $x$ and $y$ without changing the expression. Therefore from user16394's answer, $x = \frac{1}{2}(3 + \sqrt{5}), y = \frac{1}{2}(3 - \sqrt{5})$ or vice versa. Thus the value of $x^2+y^2$ is again:
$$\frac{1}{4} \left((3 + \sqrt5)^2 + (3 - \sqrt{5})^2 \right) = \frac{1}{4} \cdot 2 \left(3^2 + (\sqrt{5})^2 \right) = \boxed{7}.$$
