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(C) Let $A = (p, 0)$ and $B = (0, q)$ be the points where the line intersects the $x$-axis and $y$-axis respectively.Let $P(h, k)$ be the midpoint of

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$x^2+y^2=\frac{a^2}{4}$

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(C) Let $A = (p, 0)$ and $B = (0, q)$ be the points where the line intersects the $x$-axis and $y$-axis respectively.Let $P(h, k)$ be the midpoint of the line segment $\overline{AB}$.Given that the length of the segment $\overline{AB} = a$.Since $P(h, k)$ is the midpoint of $\overline{AB}$,we have:$h = \frac{p+0}{2} \implies p = 2h$$k = \frac{0+q}{2} \implies q = 2k$The length of the segment $\overline{AB}$ is given by the distance formula:$\sqrt{(p-0)^2 + (0-q)^2} = a$$\sqrt{p^2 + q^2} = a$Substituting $p = 2h$ and $q = 2k$ into the equation:$\sqrt{(2h)^2 + (2k)^2} = a$$\sqrt{4h^2 + 4k^2} = a$Squaring both sides:$4h^2 + 4k^2 = a^2$$h^2 + k^2 = \frac{a^2}{4}$Replacing $(h, k)$ with $(x, y)$,the locus of the midpoint is:$x^2 + y^2 = \frac{a^2}{4}$Thus,the correct option is $C$.

$A$ line moves such that the portion of it intercepted between the coordinate axes is of constant length $a$. Then,the locus of the midpoint of that line segment is

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