If a line AB of length r moves so that A and | vedclass

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(C) Given that $A$ lies on the $X$-axis,let $A \equiv (a, 0)$.Given that $B$ lies on the line $y=6x$,let $B \equiv (c, 6c)$.Let $C(h, k)$ be the mid-p

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$(x-y/3)^2+y^2=\frac{r^2}{4}$

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(C) Given that $A$ lies on the $X$-axis,let $A \equiv (a, 0)$.Given that $B$ lies on the line $y=6x$,let $B \equiv (c, 6c)$.Let $C(h, k)$ be the mid-point of $AB$.Then,$h = \frac{a+c}{2}$ and $k = \frac{0+6c}{2} = 3c$.From $k = 3c$,we get $c = \frac{k}{3}$.Substituting $c$ in the expression for $h$: $h = \frac{a + k/3}{2}$ $\Rightarrow 2h = a + \frac{k}{3}$ $\Rightarrow a = 2h - \frac{k}{3}$.Given the length $AB = r$,we have $(AB)^2 = r^2$.$(a-c)^2 + (0-6c)^2 = r^2$Substituting $a = 2h - k/3$ and $c = k/3$:$(2h - k/3 - k/3)^2 + (6(k/3))^2 = r^2$$(2h - 2k/3)^2 + (2k)^2 = r^2$$4(h - k/3)^2 + 4k^2 = r^2$$(h - k/3)^2 + k^2 = \frac{r^2}{4}$Replacing $(h, k)$ with $(x, y)$,the locus is $(x - y/3)^2 + y^2 = \frac{r^2}{4}$.

If a line $AB$ of length $r$ moves so that $A$ and $B$ always lie respectively on the $X$-axis and the line $y=6x$,then the locus of the mid-point of $AB$ is:

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