Inspired by Ethan

Let us analyze the first two circles:

For easy naming, let us call the circles $c_1,c_2,c_3,\ldots$ . The radius of these circles will be $r_1,r_2,r_3,\ldots$ and the area of these circles will be $a_1,a_2,a_3,\ldots$

First of all, since $OBQD$ is a quadrant, we know that $\angle AOC=90^{\circ}$ . Lines $OCD$ and $OAB$ are tangent to $c_2$ , so we also know that $\angle OCP = \angle OAP = \angle APC =90^{\circ}$ . We know that $AP=CP=r_2$ , therefore it is obvious that $OAPC$ is a square. With this, we can find $OP$ .

$OP = \sqrt{CP^2+AP^2} = \sqrt{r_2^2+r_2^2} = \sqrt{2r_2^2} = \sqrt{2}r_2$

Note that $PQ = r_2$ and $OQ = r_1=10$ . We can then say that

$OQ = OP + PQ\\ r_1 = \sqrt{2}r_2+r_2\\ r_2 = \dfrac{r_1}{\sqrt{2}+1} = \dfrac{10}{\sqrt{2}+1}$

Now, notice that for the circles $c_2$ and $c_3$ , $c_3$ and $c_4$ ...., the pattern repeats in the exact same manner.

This implies that $r_{n+1} = \dfrac{r_n}{\sqrt{2}+1}$

The radius of the circles, $r_1,r_2,r_3,r_4,\ldots$ are $10,\dfrac{10}{\sqrt{2}+1},\dfrac{10}{(\sqrt{2}+1)^2},\dfrac{10}{(\sqrt{2}+1)^3},\ldots$

The areas of the circles, $a_1,a_2,a_3,a_4,\ldots$ are (use $a=\pi r^2$ ): $100\pi,\dfrac{100\pi}{(\sqrt{2}+1)^2},\dfrac{100\pi}{(\sqrt{2}+1)^4},\dfrac{100\pi}{(\sqrt{2}+1)^6},\ldots$

Notice that the areas of these circles form a geometric progression where $a=100\pi$ and $r=\dfrac{1}{(\sqrt{2}+1)^2}$

Now, we are looking for the sum of the areas of all the circles when this pattern is repeated infinitely. In other words, the sum of the GP above when $n=\infty$ . Use the formula $S_{\infty} = \dfrac{a}{1-r}$ :

$A = a_1+a_2+a_3+a_4+\ldots\\ =\dfrac{100\pi}{1-\frac{1}{(\sqrt{2}+1)^2}}\\ =\dfrac{100\pi}{1-\frac{1}{3+2\sqrt{2}}}\\ =\dfrac{100\pi}{\frac{3+2\sqrt{2}-1}{3+2\sqrt{2}}}\\ =\dfrac{100\pi(3+2\sqrt{2})}{2+2\sqrt{2}}\\ =\dfrac{100\pi(3+2\sqrt{2})(\sqrt{2}-1)}{2(\sqrt{2}+1)(\sqrt{2}-1)}\\ =\dfrac{50\pi(3\sqrt{2}+4-3-2\sqrt{2})}{2-1}\\ =50\pi(\sqrt{2}+1)$

$\implies a=50,\;b=2,\;c=1,\;a+b+c=50+2+1=\boxed{53}$

Let the largest radius be $r_0 = 10$ , the next largest be $r_1$ , then $r_2$ and so on. Then we note that:

$\begin{aligned} r_0 & = r_1 + \sqrt 2 r_1 \\ \implies r_1 & = \frac {r_0}{1+\sqrt 2} \\ r_2 & = \frac {r_0}{(1+\sqrt 2)^2} \\ r_3 & = \frac {r_0}{(1+\sqrt 2)^3} \\ ... & \quad \ ... \\ \implies r_k & = \frac {r_0}{(1+\sqrt 2)^k} \end{aligned}$

The area of all the circles is given by:

$\begin{aligned} A & = \sum_{k=0}^\infty \pi r_k^2 \\ & = \pi r_0^2 \sum_{k=0}^\infty \left(\frac 1{1+\sqrt 2} \right)^{2k} \\ & = \pi r_0^2 \sum_{k=0}^\infty \left(\frac 1{3+2\sqrt 2} \right)^{k} \\ & = \pi r_0^2 \left(\frac 1{1-\frac 1{3+2\sqrt 2}} \right) \\ & = 100 \pi \left(\frac {3+2\sqrt 2}{2+2\sqrt 2} \right) \\ & = 50 \pi \left(\frac {3+2\sqrt 2}{1+\sqrt 2} \right) \\ & = 50 \pi \left(\frac {(3+2\sqrt 2)(\sqrt 2 -1)}{(1+\sqrt 2)(\sqrt 2 -1)} \right) \\ & = 50 \pi (1+\sqrt 2) \end{aligned}$

$\implies a + b + c = 50 + 2 + 1 = \boxed{53}$

The relation between radius of bigger circle ( $x$ ) and and radius of circle inscribed ( $r$ ) is: $r=\dfrac{x}{1+\sqrt{2}}.$

Radius of first circle= $10$ [ $\text{Given}$ ]

Radius of next circle= $\dfrac{10}{1+\sqrt{2}}$

Radius of next circle= $\dfrac{\frac{10}{1+\sqrt{2}}}{1+\sqrt{2}}=\dfrac{10}{(1+\sqrt{2})^2}$

$\cdot \cdot \cdot$ $\cdot \cdot \cdot$

As we know area of circle is: $\pi \color{#20A900}{r^2}$

Area of all circles: $\Rightarrow \pi\color{#20A900}{(10)^2}+\pi \color{#20A900}{\left(\dfrac{10}{1+\sqrt{2}}\right)^2}+\pi \color{#20A900}{\left(\dfrac{10}{(1+\sqrt{2})^2}\right)^2}+\pi \color{#20A900}{\left(\dfrac{10}{(1+\sqrt{2})^4}\right)^2}+\cdot \cdot \cdot \cdot$

$\implies \pi\left[100+100\left(\color{#EC7300}{\dfrac{1}{(1+\sqrt{2})^2}+\dfrac{1}{(1+\sqrt{2})^4}+\dfrac{1}{(1+\sqrt{2})^6}+\cdot \cdot \cdot \cdot } \right)\right]$

Using Geometric progression.

$\implies \pi\left[100+100\left(\color{#EC7300}{\dfrac{1}{(1+\sqrt{2})^2-1}}\right)\right]$

$\implies \pi\left[100+100\left(\dfrac{1}{2\sqrt{2}+2}\right)\right]$

$\implies \pi [100+50(\sqrt{2}-1)]$

$\implies 50\pi(1+\sqrt{2})$

$\therefore a+b+c=50+2+1=\boxed{53}$