If a line $A B$ of length $r$ moves so that $A$ and $B$ always lie respectively on $X$-axis and $y=6 x$,…
- $y=12 x$
- $(x-y / 3)^2+y^2=\frac{r^2}{2}$
- $(x-y / 3)^2+y^2=\frac{r^2}{4}$
- $y=6 x$
Solution

$C$ is the mid-point of $A B$. Let $A \equiv(a, 0), B \equiv(c, 6 c)$ and $C \equiv(h, k)$. Now, $h=\frac{a+c}{2}$ and $\frac{0+6 c}{2}=k$ $\begin{array}{ll}\Rightarrow & c=\left(\frac{k}{3}\right) \Rightarrow 2 h=a+\frac{k}{3} \\ \Rightarrow & a=\left(2 h-\frac{k}{3}\right) \\ \because & A B=r\end{array}$ $\begin{aligned} & \Rightarrow \quad(A B)^2=r^2 \Rightarrow(a-c)^2+(6 c)^2=r^2 \\ & \Rightarrow \quad\left(2 h-\frac{k}{3}-\frac{k}{3}\right)^2+\left\{\left(\frac{6 k}{3}\right)\right\}^2=r^2 \\ & \Rightarrow \quad\left(2 h-\frac{2 k}{3}\right)^2+(2 k)^2=r^2 \\ & \Rightarrow\left(h-\frac{k}{3}\right)^2+k^2=\frac{r^2}{4}\end{aligned}$ $\therefore$ Locus of $(h, k)$ is given by $\left(x-\frac{y}{3}\right)^2+y^2=\frac{r^2}{4}$
Asked in: AP EAMCET 2022 (08 Jul Shift 2)