We will prove that for each integer $L > 0$ there exists a square of side-length $L$ with side parallel to the axes of ccordinates such that any point in the interior of the square or on its boundary is invisible. This will lead to the desired result since, for any given $R$, we may choose $L$ sufficientely large so that the square covers a disk with radius $R$. Clearly, the lattice point $M(a,b)$ is invisible if and only if $\gcd (a,b) > 1$. Now, let $L > 0$ be given. Let $p_{i,j}$ be pairwise distinct primes for $i=0, \cdots, L$ and $j = 0 , \cdots , L$. From the chinese remainders theorem, we know that there exists $x,y > 0$ such that $x =- i$ modmath=inline]$p_{i,j}$[/math and $y=-j$ modmath=inline]$p_{i,j}$[/math, for all $i,j$. Then we choose the point $M(x,y)$ to be the lower left corner of our desired square, and wed are done.