<s>[INST] Calculate the maximum value of $\lfloor x\lfloor x \rfloor \rfloor + \lfloor y\lfloor y \rfloor \rfloor$. 
Given negatives $x$ and $y$ such that $\left(\dfrac{x}{2} - 1\right)^2 + \left(\dfrac{y}{2} - 1\right)^2 \le \dfrac{125}{2}$. Calculate the maximum value of $$\large \lfloor x \lfloor x \rfloor \rfloor + \lfloor y \lfloor y \rfloor \rfloor$$

We could solve for the maximum value of $x^2 + y^2$.
We have that $$\left(\frac{x}{2} - 1\right)^2 + \left(\frac{y}{2} - 1\right)^2 \le \frac{125}{2} \iff \frac{x^2 + y^2}{4} - (x + y) + 2 \le \frac{125}{2}$$
$$\iff x^2 + y^2 \le 2[121 - 2(x + y)]$$
Moreover, $$\frac{(x + y)^2}{8} - (x + y) - \frac{121}{2} \le 0 \implies x + y \in (4 - 10\sqrt5, 0)$$ since $x, y < 0$.
$$\implies x^2 + y^2 \le 2[121 - 2(4 - 10\sqrt5)] = 2(113 + 20\sqrt5)$$
But I'm uncertain about the case for $x \lfloor x \rfloor + y \lfloor y \rfloor$ or even $\lfloor x \lfloor x \rfloor \rfloor + \lfloor y \lfloor y \rfloor \rfloor$.
 [/INST] Well, we have the following problem:
$$\left(\frac{x}{\Delta}+\alpha\right)^2+\left(\frac{\text{y}}{\Delta}+\alpha\right)^2\le\frac{\rho}{\Delta}\tag1$$
Where all the constants are real numbers.
Now, if we multiply the LHS and the RHS with $\Delta^2$ we get:
$$\Delta^2\cdot\left(\left(\frac{x}{\Delta}+\alpha\right)^2+\left(\frac{\text{y}}{\Delta}+\alpha\right)^2\right)\le\Delta^2\cdot\frac{\rho}{\Delta}\tag2$$
Which we can rewrite as:
$$\left(x+\Delta\alpha\right)^2+\left(\text{y}+\Delta\alpha\right)^2\le\Delta\rho\tag3$$
This represents a circle with centre coordinates $\left(-\alpha,-\alpha\right)$ and radius $\sqrt{\Delta\rho}$.
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