Let M be the maximum number of unit disks (disks of radius 1) that can be placed inside a disk of radius 10 Let M be the maximum number of unit disks (disks of radius 1) that can be placed
inside a disk of radius 10 so that each unit disk lies entirely within the larger disk and
no two unit disks overlap.Prove that $M \ge 55$
The solution says  A $14*14$ square can be inscribed that holds 49 unit disks To this you can add (at
least) 4 more disks on each side, so at least 65 unit disks fit inside the big circle. so,$M \ge 55$.
But I cannot understand the solution
 A: As a supplement to the previous answer, let's consider the question in the first comment: how can we guarantee that we can add 4 more circles on each side? 
Take the segment of the circle that's defined by the circle $x^2 + y^2 = 10^2$ and the horizontal line at the top of the square $y = 7$. At the center of that segment, on the vertical line $x = 0$, the distance between the northernmost point on the circle, $(0, 10)$, and the point on the square directly below it, $(0, 7)$, is $10 - 7 = 3$. As we go off to the side, in order to be able to fit a circle of diameter $2$, we need to have the vertical distance between the circle and the square be greater than that diameter: $y - 7 \ge 2$. The question is how far along x we need to go before this becomes false. Since the point on the circle satisfies the equation
$$ x^2 + y^2 = 10^2 \implies y = \sqrt{10^2 - x^2}$$
we must have
$$ y - 7 \ge 2 \implies \sqrt{10^2 - x^2} - 7 \ge 2 \implies x^2 \le 100 - 81 = 19$$
so $|x| \le 4.35...$. Therefore as long as x is between $-4.35$ and $4.35$, the vertical distance is greater than 2, so a circle will fit. That gives us a space of width $8.7 \gt 4 \times 2$ so it will certainly accommodate 4 circles.
Edit: That's assuming that they are in a straight line touching the square. As David K. points out in a comment on the question, it's easy to add at least one more: just put one in the center and shift the two on each side up until they touch the circle, leaving enough space to add at least two more. That gets you to 69, which as Jack D'Aurizio's answer points out is still far from optimal.
A: In a circle with radius $10$ there is enough space to fit at least $\color{red}{76}$ disjoint unit circles:

It is much more challenging to prove that we cannot fit $\geq 92$ disjoint unit circles in a circle with radius $10$. According to this reference, in a disk with radius $10$ we may fit at most $80$ disjoint circles with radius $1$, so the configuration above is not very far from being optimal.
A: A unit disk has diameter 2 so can be inscribed in $2 \times 2$ square. A $14 \times 14$ square this holds $7 \times 7 = 49$ of $2 \times 2$ squares and hence at least 49 circles.
It is easy to see when you inscribe a maximum square of side $s$ into the circle, its half-diagonal is exactly the radius of the circle, and hence $r = s/\sqrt{2}$. Here, $s = r\sqrt{2} = 10 \sqrt{2} > 14.1$ so a square of size $14$ can certainly be inscribed.
Adding 4 more disks on each side creates $49+4\times 4 = 65$ circles.
