**Problem:** Let \({C_n}\) be an infinite sequence of circles lying in the positive quadrant of the \(XY\)-plane, with strictly decreasing radii and satisfying the following conditions. Each \(C_n\) touches both the \(X\)-axis and the \(Y\)-axis. Further, for all \(n\geq 1\), the circle \(C_{n+1}\) touches the circle \(C_n\) externally. If \(C_1\) has radius \(10\: cm\), then show that the sum of the areas of all these circles is \(\frac{25\pi}{3\sqrt{2}-4} \: cm^2\).

**Solution:** Consider the following diagram where the Green line segment is \(R_n\), the radius of the \(n^{th}\) circle, and the Yellow line segment is \(R_{n+1}\).

As we are told about the symmetricity of the figure in the problem we can say that:

\(\sqrt{2}R_{n+1} + R_{n+1} + R_n = \sqrt{2} R_n\)

\(=> R_{n+1}(\sqrt{2}+1)=R_n(\sqrt{2}-1)\)

\(=> R_{n+1}= (3-2\sqrt{2})R_n\)

Let’s say \(=> R_{n+1}= \alpha.R_n\).

Now the total sum of the areas of the circles is:

\((\pi R_1^2 + \pi R_2^2 + \cdots ) = \pi (R_1^2 + R_2^2 + R_3^2 + \cdots )\)

Now as \(R_{n+1}= \alpha.R_n\), we can say that:

\(\pi (R_1^2 + R_2^2 + R_3^2 + \cdots ) = \pi (R_1^2 + \alpha^2 R_1^2 + \alpha^4 R_1^2 + \cdots ) = \pi \frac{R_1^2}{1-\alpha^2}\) as \(\alpha^2 < 1\).

Substituting the value of \(\alpha = 3-2\sqrt{2}\) and \(R_1 = 10 \: cm\)we have,

Sum = \(\frac{25\pi}{3\sqrt{2}-4} \: cm^2\).

Hence Proved.

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