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(D) Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$. Since it passes through $(0,0)$ and $(1,0)$,the radius $r$ is the distance from $(h,k

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$9$

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(D) Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$. Since it passes through $(0,0)$ and $(1,0)$,the radius $r$ is the distance from $(h,k)$ to $(0,0)$,so $r^2 = h^2+k^2$.Expanding the equation: $x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = h^2+k^2$,which simplifies to $x^2+y^2-2hx-2ky=0$.Since the circle passes through $(1,0)$,we substitute $x=1, y=0$: $1^2+0^2-2h(1)-2k(0)=0$,which gives $1-2h=0$,so $h=1/2$.The circle touches $x^2+y^2=9$ (a circle with center $(0,0)$ and radius $3$). The distance between the centers is equal to the difference of the radii (since the smaller circle is inside the larger one): $\sqrt{h^2+k^2} = 3 - r = 3 - \sqrt{h^2+k^2}$.Thus,$2\sqrt{h^2+k^2} = 3$,so $\sqrt{h^2+k^2} = 3/2$.Squaring both sides,$h^2+k^2 = 9/4$.Therefore,$4(h^2+k^2) = 4(9/4) = 9$.

Let the centre of a circle,passing through the points $(0,0)$ and $(1,0)$ and touching the circle $x^2+y^2=9$,be $(h, k)$. Then for all possible values of the coordinates of the centre $(h, k)$,$4(h^2+k^2)$ is equal to .............

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