(Level VI, part II)Geometricum geometriorum (HP III, Prisoner of "the circle")

I'm going to prove the general case. (In this case, $a = 5$ and $b = 4$ ).

Suppose WLOG, $O = (0,0)$ , $B = (b,0)$ and $P_j = (a \cos \left((j - 1) \frac{2\pi}{n} \right), a \sin \left((j - 1) \frac{2\pi}{n} \right))$ for $j = 1, 2, ... , n.$

The centers $O_j$ of the tangent circumferences will be in the line $OP_j$ , i.e, $O_j = (a \lambda_j \cos\left((j - 1) \frac{2\pi}{n} \right), a \lambda_j \sin \left((j - 1) \frac{2\pi}{n} \right))$ Because of $d(O_j, B) = d(O_j, P_j)$ we have $\left(a \lambda_j \cos \left((j - 1) \frac{2\pi}{n} \right) - b \right)^2 + a^2 \lambda_j^2 \sin^2 \left((j - 1) \frac{2\pi}{n} \right) = a^2(1 - \lambda_j)^2 \left(\cos^2 \left((j - 1) \frac{2\pi}{n} \right) + \sin^2 \left((j - 1) \frac{2\pi}{n} \right)\right)$ ,i.e, $b^2 + a^2(\lambda_j)^2 - 2ab\lambda_j \cos \left((j - 1) \frac{2\pi}{n} \right) = a^2 - 2\lambda_j a^2 + a^2 \lambda_j^2 \Rightarrow$ $\lambda_j = \frac{a^2 - b^2}{2a\left(a - b \cos \left((j - 1) \frac{2\pi}{n} \right)\right)}$ The radii of the circumferencies are $r_j = d(O_j, p_j) = a(1 - \lambda_j)$ and then, as $n$ tends to infinitum, the limit of the arithmetic mean of the lengths of all circumferences tangent to the first one, passing through the points of division and a fixed point B is: $M = \displaystyle \lim_{n \to \infty} \frac{1}{n} \sum_{ j = 1}^{\infty} 2\pi a(1 - \lambda_j) = \lim_{n \to \infty} \frac{2\pi}{n} \sum_{ j = 1}^{\infty} \left[ a - \frac{a^2 - b^2}{2\left(a - b \cos \left((j - 1) \frac{2\pi}{n} \right)\right)}\right] =$ $\displaystyle = 2\pi a - \lim_{n \to \infty} \frac{\pi}{n} \sum_{ j = 1}^{\infty} \left[ \frac{a^2 - b^2}{2\left(a - b \cos \left((j - 1) \frac{2\pi}{n} \right)\right)}\right] = 2\pi a - \frac{1}{2} \int_0^{2\pi} \frac{a^2 - b^2}{a - b\cos(t)} dt =$ $= 2\pi a - \pi \sqrt{a^2 - b^2}$ Substituing $a=5 , b = 4$ we get, $M = 7 \pi$

Note.- For doing the integral, you can make the change $t = \arctan(x/2)$ . If someone wishes it, I can do it...