Gauss circle problem : a simple asymptotic estimation.

Pick a square with unit side length centered at every lattice point inside the region $x^2+y^2=n^2$, i.e. the circle with radius $n$ centered at the origin. Those squares entirely cover the circle with radius $n-\frac{1}{\sqrt{2}}$, but any square lies inside a circle with radius $n+\frac{1}{\sqrt{2}}$, hence the number of lattice points is between $\pi\left( n-\frac{1}{\sqrt{2}}\right)^2$ and $\pi\left(n+\frac{1}{\sqrt{2}}\right)^2$, so it is $\pi n^2+O(n)$.

For info about the Voronoi bound, please see this survey.

The estimate $\displaystyle \frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}} = \int_0^1 \sqrt{1-t^2}dt + o(1)$ can easily be refined .

Indeed, $$\begin{align}0\leq \frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}} - \int_0^1 \sqrt{1-t^2}dt &= \sum_{k=0}^{n-1} \int_{k/n}^{(k+1)/n} \left(\sqrt{1-\frac{k^2}{n^2}} - \sqrt{1-t^2}\right)dt\\ &\leq\sum_{k=0}^{n-1} \int_{k/n}^{(k+1)/n} \left(\sqrt{1-\frac{k^2}{n^2}} - \sqrt{1-\frac{(k+1)^2}{n^2}}\right)dt \\ &= \frac 1n \end{align}$$

Hence $\displaystyle \frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}} = \int_0^1 \sqrt{1-t^2}dt + O(\frac 1n)$ and $$1+4\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor = \pi n^2 + O(n)$$