A line segment AB of length moves such that the points A and B remain on the periphery of …

Question

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 $. Then the locus of the point, that divides the line segment AB in the ratio 2 : 3, is a circle of radius :

Accepted Answer

Given, length of $A B=\lambda$

So, $A C=\frac{\lambda}{2}$ and $A M=\frac{2 \lambda}{5}$

$$
C M=A C-A M=\frac{\lambda}{2}-\frac{2 \lambda}{5}=\frac{\lambda}{10}
$$

Now, from $\triangle O C M$, we get

$$
\[\begin{gathered} O M^2=C M^2+O C^2 \ \Rightarrow  x^2+y^2=\left(\frac{\lambda}{10}\right)^2+\left(\lambda^2-\frac{\lambda^2}{4}\right) \ \Rightarrow  x^2+y^2=\frac{\lambda^2}{100}+\frac{3 \lambda^2}{4}=\frac{\lambda^2+75 \lambda^2}{100} \ \Rightarrow  x^2+y^2=\frac{76 \lambda^2}{100}=\frac{19 \lambda^2}{25} \ \therefore  x^2+y^2=\left(\frac{\sqrt{19} \lambda}{5}\right)^2 \end{gathered}$$
\]

So, radius is $\frac{\sqrt{19} \lambda}{5}$.

Solutions · 1