A line segment AB of length λ moves such that the points A and B...

Given,

A line segment $AB$ of length $\lambda$ moves such that the points $A$ and $B$ remain on the periphery of a circle of radius $\lambda$,

Now taking the points on the circle of radius $\lambda$ as $B\left(\lambda cos{\theta}_{1}, \lambda sin{\theta}_{1}\right) \text{ and } A\left(\lambda cos{\theta}_{2}\lambda sin{\theta}_{2}\right)$ and taking the point $$P\left(h, k\right)$ which divides the line segment in $AB$ of length $\lambda$ in $2: 3$

Now plotting the diagram we get,

Now, let $O$ be the origin and radius of circle is $\lambda$ and $AB = \lambda$ and using distance formula we get,

$AB = \lambda = \sqrt{{\left(\lambda cos{\theta}_{1} - \lambda cos{\theta}_{2}\right)}^{2} + {\left(\lambda sin{\theta}_{1} - \lambda sin{\theta}_{2}\right)}^{2}}$

$⇒ 1 = 2 - 2\cos \left({\theta}_{1} - {\theta}_{2}\right)$

$⇒ \cos \left({\theta}_{1} - {\theta}_{2}\right) = \dfrac{1}{2}$

Now using section formula we get,

$h = \dfrac{2\lambda cos{\theta}_{1} + 3\lambda cos{\theta}_{2}}{5}$ and $k = \dfrac{2\lambda sin{\theta}_{1} + 3\lambda sin{\theta}_{2}}{5}$

Now squaring and adding above two value we get,

${h}^{2} + {k}^{2} = \dfrac{{\lambda}^{2}}{25}\left[4 + 9 + 12\left(\cos \left({\theta}_{1} - {\theta}_{2}\right)\right)\right]$

$⇒ {h}^{2} + {k}^{2} = \dfrac{{\lambda}^{2}}{25} \cdot 19$

Hence, Radius $= \dfrac{\lambda}{5}\sqrt{19}