This problem has appeared in one of the MBA entrance papers. I know the answer and have seen that it's possible to get an output on Desmos etc. to verify the result. But I am having a hard time figuring out a simple, elegant way to get to the answer, without a graphing tool at my disposal. It goes like,
Rekha drew a circle of radius $2$ cm on a graph paper of grid $1 cm × 1 cm$. She then calculated the area of the circle by adding up only the number of full unit-squares that fell within the perimeter of the circle. If the value that Rekha obtained was $d$ square cm. less than the correct value, then find the minimum possible value of $d$?
The correct answer is $5.56$, and it is so because Rekha can have a maximum of seven squares within her circle. Now I do not know how to prove that, as I said, "manually". I mean, how to fit that circle, beginning with what coordinate for which location, to arrive at seven full squares definitely within Rekha's circle?
A simple explanation without hard geometry, shall be highly appreciated!!
Pick a square that covers the centre of the circle. Then at most this square and its eight neighbours (forming a $3\times 3$ block) are covered by the circle. Since the diagonal of this block has length $3\sqrt 2>4$, you cannot have both the top left and the lower right square covered, nor can you have both the top right and lower left square covered. It follows that at most seven squares are covered. To see that seven can indeed be achieved, simply let the circle pass through two neighbouring square vertices. For example, the circle passes trough $(0,0)$ and $(1,0)$ if its centre is at $(\frac12,\sqrt{4-\frac14})=(\frac12,\frac12\sqrt{15})$.
You can check that $(2,3)$ is still inside the circle: $$ \left(2-\frac12\right)^2+\left(3-\frac12\sqrt{15}\right)^2 =\frac94+9-3\sqrt{15}+\frac{15}4=15-3\sqrt{15}$$ Numerically, this is $\approx 3.38$, but you can also see that it is $<4$ because that's equivalent to $3\sqrt{15}>11$, or $9\cdot 15>121$.
Likewise, $(2,1)$ is inside the circle: $$ \left(2-\frac12\right)^2+\left(1-\frac12\sqrt{15}\right)^2 =\frac94+1-\sqrt{15}+\frac{15}4=7-\sqrt{15}$$ which is $<4$ because $\sqrt{15}>3$.
Then by convexity of the cirle, $[-1,2]\times [1,3]$ is covered (=six squares) as well as $[0,1]\times[0,1]$.