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(D) The equation of the circle is $x^{2}+y^{2}-8x-4y+16=0$.Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-4, f=-2, c=16$.The center of the circle

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$\frac{64}{5}$

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(D) The equation of the circle is $x^{2}+y^{2}-8x-4y+16=0$.Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-4, f=-2, c=16$.The center of the circle is $(-g, -f) = (4, 2)$ and the radius $R = \sqrt{g^{2}+f^{2}-c} = \sqrt{16+4-16} = \sqrt{4} = 2$.The length of the tangent from the origin $(0, 0)$ to the circle is $L = \sqrt{S_{1}} = \sqrt{0^{2}+0^{2}-8(0)-4(0)+16} = \sqrt{16} = 4$.The length of the chord of contact $AB$ is given by the formula $AB = \frac{2LR}{\sqrt{L^{2}+R^{2}}}$.Substituting the values,$AB = \frac{2 	imes 4 	imes 2}{\sqrt{4^{2}+2^{2}}} = \frac{16}{\sqrt{16+4}} = \frac{16}{\sqrt{20}} = \frac{16}{2\sqrt{5}} = \frac{8}{\sqrt{5}}$.Therefore,$(AB)^{2} = \left(\frac{8}{\sqrt{5}}ight)^{2} = \frac{64}{5}$.

Let the tangents drawn from the origin to the circle $x^{2}+y^{2}-8x-4y+16=0$ touch it at the points $A$ and $B$. The $(AB)^{2}$ is equal to

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