JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Circle

Let the circle have centre $O$ and radius $\lambda$, and let $AB$ be a chord of length $\lambda$. The perpendicular from $O$ to $AB$ meets it at its midpoint $M$, where $OM=\sqrt{{\lambda }^2-(\lambda /2{)}^2}=\frac{\sqrt{3}}{2}\lambda$, a value that stays constant as the chord moves. The point $P$ dividing $AB$ in the ratio $2:3$ has $AP=\frac{2}{5}\lambda$, so $MP=\left|\frac{\lambda }{2}-\frac{2\lambda }{5}\right|=\frac{\lambda }{10}$. Since $OM\perp AB$, $O{P}^2=O{M}^2+M{P}^2=\frac{3{\lambda }^2}{4}+\frac{{\lambda }^2}{100}=\frac{76{\lambda }^2}{100}=\frac{19{\lambda }^2}{25}$. As $OP$ is the same for every position of the chord, the locus is a circle of radius $\frac{\sqrt{19}}{5}\lambda$.