A rod PQ of length 2a moves with its ends on | vedclass

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(D) Let the coordinates of $P$ be $(2x_0, 0)$ and $Q$ be $(0, 2y_0)$ such that the length $PQ = \sqrt{(2x_0)^2 + (2y_0)^2} = 2a$. This simplifies to $

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$x^2 + y^2 = a^2$

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(D) Let the coordinates of $P$ be $(2x_0, 0)$ and $Q$ be $(0, 2y_0)$ such that the length $PQ = \sqrt{(2x_0)^2 + (2y_0)^2} = 2a$. This simplifies to $4x_0^2 + 4y_0^2 = 4a^2$,or $x_0^2 + y_0^2 = a^2$. Since $\Delta OPQ$ is a right-angled triangle with the right angle at the origin $O(0,0)$,the circumcenter $M(h, k)$ is the midpoint of the hypotenuse $PQ$. Thus,$h = \frac{2x_0 + 0}{2} = x_0$ and $k = \frac{0 + 2y_0}{2} = y_0$. Substituting $x_0 = h$ and $y_0 = k$ into the equation $x_0^2 + y_0^2 = a^2$,we get $h^2 + k^2 = a^2$. Therefore,the locus of the circumcenter is $x^2 + y^2 = a^2$.

$A$ rod $PQ$ of length $2a$ moves with its ends on the coordinate axes. Find the locus of the circumcenter of $\Delta OPQ$.

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