Area of infinite circles

if we take any adjacent circle like above we will get a similar triangles

$\triangle abc \sim \triangle ACO$

$\Rightarrow \frac { \overline { AO } }{ \overline { ac } } =\frac { \overline { CO } }{ \overline { bc } }$

$\Rightarrow \frac { 9 }{ { r }_{ n }+{ r }_{ n+1 } } =\frac { 2 }{ { r }_{ n }-{ r }_{ n+1 } }$ since $\begin{cases} \overline { ac } ={ r }_{ n }+{ r }_{ n+1 } \\ \overline { bc } ={ r }_{ n }-{ r }_{ n+1 } \end{cases}$

$\Rightarrow { r }_{ n+1 }=\frac { 7 }{ 11 } { r }_{ n }$

Now

${ r }_{ 1 }=2\quad ,\quad { r }_{ 2 }=2{ \left( \frac { 7 }{ 11 } \right) }^{ 1 }\quad ,\quad { r }_{ 2 }=2{ \left( \frac { 7 }{ 11 } \right) }^{ 2 },.....,{ r }_{ n }=2{ \left( \frac { 7 }{ 11 } \right) }^{ n-1 }$

the Area of a circle ${ A }_{ n }=\pi { \left( { r }_{ n } \right) }^{ 2 }=4\pi { \left( \frac { 49 }{ 121 } \right) }^{ n-1 }$

the sum of all circles

$\sum _{ n=1 }^{ \infty }{ { A }_{ n } } =4\pi \sum _{ n=1 }^{ \infty }{ { \left( \frac { 49 }{ 121 } \right) }^{ n-1 } } =\frac { 4\pi }{ 1-\frac { 49 }{ 121 } } =\frac { 121 }{ 18 } \pi$

a+b = 121 + 18 = 139

another Solution

$\triangle APR \sim \triangle ASN$

so the Area of the shaded Area is similar

$\Rightarrow \frac { area\quad of\quad circle\quad in\quad \triangle APR }{ area\quad of\quad circle\quad in\quad \triangle ANS } =\left ({\frac {AJ}{AG}} \right )^{2}$

let the area of circles in $\triangle ANS$ = $\alpha$

so the area of circles in $\triangle APR$ = $\alpha -4\pi$

$\Rightarrow \frac { \alpha -4\pi }{ \alpha } =\frac { { 7 }^{ 2 } }{ { 11 }^{ 2 } } \Rightarrow \alpha =\frac { 121 }{ 18 } \pi$

a+b = 121 + 18 = 139

In my diagram above $\triangle AOD \sim \triangle APQ$ . Then we have the following relation:

$\displaystyle{\frac{r_1}{r_2}=\frac{\overline{AO}}{\overline{AP}}}$

$\displaystyle{\frac{2}{r_2}=\frac{9}{9-2-r_2}}$

Solving for $r_2$ :

$\displaystyle{\frac{2}{r_2}=\frac{9}{7-r_2}}$

$\displaystyle{9r_2=14-2r_2}$

$\displaystyle{11r_2=14}$

$\displaystyle{r_2=\frac{14}{11}}$

Since all the circles are tangent two by two, and they are also tangent to the segment $\overline{AD}$ , all the radii of adjacent circles starting from $r_1$ are decreasing in the same ratio. This ratio is $\displaystyle{\frac{\frac{14}{11}}{2}=\frac{7}{11}}$ .

The ratio of the areas of two circles is the same as the ratio of their radii squared. Then the area of echa circle starting from the circle with radius $r_1$ is decreasing in the ratio of $\displaystyle{\frac{49}{121}}$ .

The sum of the areas of all circles will be the sum of the infinite terms of an Geometric Progression with first term $4\pi$ and common ratio $\frac{49}{121}$ . The sum will be:

$\displaystyle{S=\frac{4 \pi}{1-\frac{49}{121}}}$

$\displaystyle{S=\frac{4 \pi}{\frac{72}{121}}}$

$\displaystyle{S=\frac{484 \pi}{72}=\frac{121 \pi}{18}}$

Then the sum is $\boxed{\frac{121}{18}\pi}$ , and the answer for this question is $\boxed{a+b=121+18=139}$