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(D) Let the line be $\frac{x}{a} + \frac{y}{b} = 1$. Since it is at a constant distance $c$ from the origin $(0,0)$,we have $\frac{1}{\sqrt{\frac{1}{a

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$x^2+y^2=4c^2$

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(D) Let the line be $\frac{x}{a} + \frac{y}{b} = 1$. Since it is at a constant distance $c$ from the origin $(0,0)$,we have $\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} = c$,which implies $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$.Points $A$ and $B$ are $(a, 0)$ and $(0, b)$ respectively.The circle passing through $O(0,0)$,$A(a,0)$,and $B(0,b)$ has the equation $x^2 + y^2 - ax - by = 0$.The centre of this circle is $(h, k) = (\frac{a}{2}, \frac{b}{2})$.Thus,$a = 2h$ and $b = 2k$.Substituting these into the line equation condition: $\frac{1}{(2h)^2} + \frac{1}{(2k)^2} = \frac{1}{c^2}$.$\frac{1}{4h^2} + \frac{1}{4k^2} = \frac{1}{c^2} \Rightarrow \frac{h^2 + k^2}{4h^2k^2} = \frac{1}{c^2}$.This approach is complex; alternatively,since the triangle $OAB$ is a right-angled triangle at $O$,the hypotenuse $AB$ is the diameter of the circle.The centre is the midpoint of $AB$,which is $(\frac{a}{2}, \frac{b}{2})$.The distance from origin to line $AB$ is $c = \frac{|ab|}{\sqrt{a^2+b^2}}$.Let the centre be $(x, y) = (\frac{a}{2}, \frac{b}{2})$,so $a=2x, b=2y$.$c = \frac{|(2x)(2y)|}{\sqrt{(2x)^2+(2y)^2}} = \frac{4|xy|}{2\sqrt{x^2+y^2}} = \frac{2|xy|}{\sqrt{x^2+y^2}}$.Squaring both sides: $c^2 = \frac{4x^2y^2}{x^2+y^2}$.Actually,the locus of the midpoint of the hypotenuse of a right triangle with fixed distance $c$ from the vertex is $x^{-2} + y^{-2} = c^{-2}$ if the axes are the legs. However,given the options,let's re-evaluate: The distance from origin to line $x/a + y/b = 1$ is $c = \frac{1}{\sqrt{1/a^2 + 1/b^2}}$. The centre is $(a/2, b/2)$.If the question implies the circle passes through $O, A, B$,the diameter is $AB$. The distance from origin to $AB$ is $c$. The locus of the midpoint of $AB$ where $AB$ is a line at distance $c$ from origin is $x^{-2} + y^{-2} = c^{-2}$.

$A$ line is at a constant distance $c$ from the origin and meets the coordinate axes in $A$ and $B$. The locus of the centre of the circle passing through $O, A, B$ is

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