Lattice Points

There's no easy way to do this; at least A000328 doesn't list any closed-form solution.

By counting patiently, we obtain the answer $\boxed{317}$ . Or otherwise just go to the OEIS link above and take the term $n=10$ (the 11th term, as it starts with $n=0$ ).

When I did this, I counted in a slightly easier way than counting every point. We can divide the circle into four quarters, each quarter holding one quadrant plus one half-axis. I take one quarter that holds the positive x-axis along with the first quadrant; the remaining quarters follow (the second quarter holds the positive y-axis along with second quadrant, the third quarter holds the negative x-axis along with third quadrant, the fourth quarter holds the negative y-axis along with fourth quadrant). Thus if there are $n$ lattice points in one quarter, the answer is $4n+1$ as we have $4$ quarters each containing $n$ points and $1$ center point $(0,0)$ that doesn't belong to any quarter.

Now, observe that in the quarter we choose, all lattice points are in the form of $(x,y)$ for $x \ge 0, y \ge 1$ . Thus, if we know that all lattice points on line $y=c$ lies on or to the left of $x=d$ , then we can easily compute the number of such points as $\lfloor d \rfloor$ . Additionally, we know the circle intersects $y=c$ on the line $x=\sqrt{100-c^2}$ by Pythagorean theorem, thus basically we're summing $\sum_{c=0}^{\infty} \lfloor \sqrt{100-c^2} \rfloor$ , where the upper bound is as long as the square root is defined. Thus we obtain $n = 10+9+9+9+9+8+8+7+6+4+0 = 79$ by careful counting, and consequently we obtain $4n+1 = 317$ as written above.

Using brute force in Python:

t=set({})   
for x in range(-10,11):
        for y in range(-10,11):
        if x**2+y**2<=100:
            t.add((x,y));
len(t);

Given a disc of radius $r$ , the number $N(r)$ of lattice points inside the disk can be evaluated counting the number of lattice point on the $x$ and $y$ axis, which is $4r+1$ and the number of points inside one of the quadrant multiplied by $4$ . Now, considering $1 \leq k \leq r-1$ , it's easy to see that the number of points in a quadrant is

$\displaystyle \sum_{k=1}^{r-1} \lfloor \sqrt{r^2-k^2} \rfloor$ .

Hence,

$\displaystyle N(r)=4r+1+4\sum_{k=1}^{r-1} \lfloor \sqrt{r^2-k^2} \rfloor=1+4\Bigg[\sum_{k=1}^{r-1} \lfloor \sqrt{r^2-k^2} \rfloor+r \Bigg]$ .

When $r=10$ we get

$\displaystyle N(r=10)=\boxed{317}$