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If a line $A B$ of length $r$ moves so that $A$ and $B$ always lie respectively on $X$-axis and $y=6 x$, then the locus of mid-point of $A B$ is
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The correct answer is:
$(x-y / 3)^2+y^2=\frac{r^2}{4}$
Given, $A$ lies on $X$-axis and $B$ lies on $y=6 x$ and $A B=r$.
$C$ 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
$\left(x-\frac{y}{3}\right)^2+y^2=\frac{r^2}{4}$
$C$ 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
$\left(x-\frac{y}{3}\right)^2+y^2=\frac{r^2}{4}$
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