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(D) Let the line be $\frac{x}{a} + \frac{y}{b} = 1$. Since the distance from the origin $(0,0)$ to the line is $c$,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 the distance from the origin $(0,0)$ to the line is $c$,we have $\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} = c$,which implies $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$.Since the circle passes through $O(0,0)$,$A(a,0)$,and $B(0,b)$,the circle is $x^2 + y^2 - ax - by = 0$.The centre of this circle is $(h, k) = (\frac{a}{2}, \frac{b}{2})$.Thus,$a = 2h$ and $b = 2k$.Substituting these into the equation $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$,we get $\frac{1}{(2h)^2} + \frac{1}{(2k)^2} = \frac{1}{c^2}$.$\frac{1}{4h^2} + \frac{1}{4k^2} = \frac{1}{c^2} \Rightarrow \frac{h^2+k^2}{4h^2k^2} = \frac{1}{c^2}$.Wait,the standard approach for the locus of the midpoint of the hypotenuse of a right triangle with fixed legs is different. Let the centre be $(x, y)$. Since the triangle $OAB$ is a right-angled triangle at $O$,the centre of the circle is the midpoint of the hypotenuse $AB$. Thus $A = (2x, 0)$ and $B = (0, 2y)$.The line equation is $\frac{X}{2x} + \frac{Y}{2y} = 1$. The distance from origin is $c = \frac{|-1|}{\sqrt{(\frac{1}{2x})^2 + (\frac{1}{2y})^2}} = \frac{1}{\sqrt{\frac{1}{4x^2} + \frac{1}{4y^2}}}$.$c^2 = \frac{1}{\frac{1}{4x^2} + \frac{1}{4y^2}} = \frac{4x^2y^2}{x^2+y^2}$. This does not match the options. Let's re-evaluate.Actually,the distance from origin to line $Ax+By+C=0$ is $c = \frac{|C|}{\sqrt{A^2+B^2}}$. For $\frac{x}{a} + \frac{y}{b} = 1$,$c = \frac{1}{\sqrt{1/a^2+1/b^2}}$. The centre is $(a/2, b/2)$. The locus is $x^2+y^2 = (a/2)^2 + (b/2)^2 = \frac{a^2+b^2}{4}$. Using $1/c^2 = 1/a^2 + 1/b^2 = \frac{a^2+b^2}{a^2b^2}$,we have $a^2+b^2 = \frac{a^2b^2}{c^2}$.Given the options,the intended answer is $x^2+y^2 = c^2$ is incorrect. Let's check $x^2+y^2 = c^2/4$ or similar. Actually,if $c$ is the distance,the locus is $1/x^2 + 1/y^2 = 4/c^2$.

$A$ 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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