# Circle Area Geometric Series

Here is a question I was asked recently (click for zoomed-in version). Let's look at a case where the first circle has a radius of $1$. It is a scaled version of the problem under consideration. The center of the big circle is at $(1,1)$, the center of the little circle is at $(a,a)$, and the point of contact is $\Big(1 - \frac{1}{\sqrt{2}},1 - \frac{1}{\sqrt{2}} \Big)$. We know that the radius of the small circle is equal to the distance from the center of the small circle to the point of contact.

$a^2 = \Big(a - 1 + \frac{1}{\sqrt{2}} \Big)^2 + \Big(a - 1 + \frac{1}{\sqrt{2}} \Big)^2$

Solving for $a$ yields $a = 3 - 2 \sqrt{2}$. The ratio of the area of the small circle to the big circle is $a^2 = 17 - 12 \sqrt{2}$.

Now back to the problem at hand. The area of the circle of radius $10$ is $100 \pi$. The remaining circles follow in an infinite geometric series with common ratio $17 - 12 \sqrt{2}$. The total area is therefore:

$A = \frac{100 \pi}{1 - (17 - 12 \sqrt{2})} = \frac{100 \pi}{12 \sqrt{2} - 16} = \frac{25 \pi}{3 \sqrt{2} - 4}$ Note by Steven Chase
1 year, 1 month ago

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@Steven Chase Thanks you so much sir.

- 1 year, 1 month ago

@Steven Chase BTW how do you predicted that they will follow geometric progression?

- 1 year, 1 month ago

Because the small circle just becomes the new big circle, and the whole thing repeats again

- 1 year, 1 month ago

Clever deduction. Nice!

- 1 year, 1 month ago