Circles in circle

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If we are given one big circle and infinite amount of smaller circles with equal radius (of course radius of the smaller is < radius of the big one) and we have to put in the center of the big circle one small,and from then we have to fill in the big circle with smaller circles.No overlapping or out of bounds is allowed.What will be the total number of circles inside the big circle odd or even? I think it can be both.What do you think?You put finite circles in the big one

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circles

Upper bound for the number $k$ of small circles (including the central one): $$ k \le \left\lfloor \frac{r_b^2}{r_s^2} \right\rfloor $$

Non-overlapping of circle $i$ and circle $j$, both with radius $r_s$ and center $(X_i, Y_i)$ and $(X_j, Y_j)$: $$ (X_i - X_j)^2 + (Y_i - Y_j)^2 \ge 4 r_s^2 \quad (C1) $$ where we include the inner small circle with $(X_0, Y_0) = (0,0)$: $$ X_i^2 + Y_i^2 \ge 4 r_s^2 $$

Staying within the big circle: $$ X_i^2 + Y_i^2 \le (r_b - r_s)^2 \quad (C2) $$

If $r_b - r_s < 2 r_s$ only the central circle fits in. My guess is that for any given $k$ and $r_s$ we can find a $r_b$ such that $k$ spheres fit in and $k+1$ do not.

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The question is linked to Close-packing of equal spheres. In 2d plane it is hexagon package so each package will have odd number of circles since each new layer has even number of circles + 1 in the center.

EDIT: Ok. This is wrong answer.

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The requirement that one of the small circles has to be exactly centered in the large one means that this is not exactly the usual circle packing in a circle problem.

However, the solution to the usual problem implies the answer to your question, because the known optimal solutions for packing 7 and 8 small circles in a larger circle do have one of the small circles at the center (and the rest hugging the periphery of the large circle).

Thus for diameter ratios of $3$ or $3.5$ the total number of small circles will be $7$ and $8$, respectively (see the table in the linked Wikipedia article), so both odd and even outcomes are possible.