If one of the diameters of the circle x^2+y^2 | vedclass

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(D) The given circle is $x^2+y^2-2x-6y+6=0$.Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=-3, c=6$.The center of this circle is $(-g, -f) = (1,

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$3$

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(D) The given circle is $x^2+y^2-2x-6y+6=0$.Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=-3, c=6$.The center of this circle is $(-g, -f) = (1, 3)$ and its radius $r = \sqrt{g^2+f^2-c} = \sqrt{1+9-6} = \sqrt{4} = 2$.$A$ diameter of this circle is a chord of the larger circle with center $O(2, 1)$.The length of this chord is equal to the diameter of the smaller circle,which is $2r = 2(2) = 4$.Let $A(1, 3)$ be the center of the smaller circle and $B$ be a point on the circumference of the smaller circle such that $AB$ is the radius $r=2$.The distance $OA$ between the centers $(2, 1)$ and $(1, 3)$ is $OA = \sqrt{(2-1)^2 + (1-3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1+4} = \sqrt{5}$.In the right-angled triangle formed by the center of the larger circle $O$,the center of the smaller circle $A$,and a point $B$ on the chord,we have $R^2 = OA^2 + r^2$,where $R$ is the radius of the larger circle.$R^2 = (\sqrt{5})^2 + 2^2 = 5 + 4 = 9$.Therefore,$R = \sqrt{9} = 3$.

If one of the diameters of the circle $x^2+y^2-2x-6y+6=0$ is a chord to a larger circle with center $(2,1)$,then the radius of the larger circle is:

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