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

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(B) The given circle is $x^{2}+y^{2}-2x-6y+6=0$.Comparing this with the general equation $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-1$ and $f=-3$.The center

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(B) The given circle is $x^{2}+y^{2}-2x-6y+6=0$.Comparing this with the general equation $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-1$ and $f=-3$.The center of the circle is $(-g, -f) = (1, 3)$.The radius of the circle is $R = \sqrt{g^{2}+f^{2}-c} = \sqrt{(-1)^{2}+(-3)^{2}-6} = \sqrt{1+9-6} = \sqrt{4} = 2$.Since a diameter of this circle is a chord of another circle $'C'$ with center $(2, 1)$,the radius of the circle $'C'$ (let it be $r$) forms a right-angled triangle with the distance between the centers and the radius of the first circle.The distance $d$ between the centers $(1, 3)$ and $(2, 1)$ is $d = \sqrt{(2-1)^{2}+(1-3)^{2}} = \sqrt{1^{2}+(-2)^{2}} = \sqrt{1+4} = \sqrt{5}$.In the right-angled triangle formed,the hypotenuse is the radius $r$ of circle $'C'$,and the other two sides are the distance $d$ and the radius $R$ of the first circle.Thus,$r^{2} = d^{2} + R^{2} = (\sqrt{5})^{2} + (2)^{2} = 5 + 4 = 9$.Therefore,$r = 3$.

If one of the diameters of the circle $x^{2}+y^{2}-2x-6y+6=0$ is a chord of another circle $'C'$,whose center is at $(2,1)$,then its radius is..........

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