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

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(A) Given circle equation: $x^{2}+y^{2}-4x-6y+9=0$. Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-2, f=-3, c=9$. Centre $B = (-g, -f) = (2, 3)$.

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$3$

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(A) Given circle equation: $x^{2}+y^{2}-4x-6y+9=0$. Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-2, f=-3, c=9$. Centre $B = (-g, -f) = (2, 3)$. Radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{(-2)^{2}+(-3)^{2}-9} = \sqrt{4+9-9} = 2$. Let the centre of the required circle be $A(1, 1)$. The distance $AB$ between the centres is $AB = \sqrt{(2-1)^{2}+(3-1)^{2}} = \sqrt{1^{2}+2^{2}} = \sqrt{5}$. Since the diameter of the first circle is a chord of the second circle,the radius $R$ of the second circle is the hypotenuse of the right-angled triangle formed by the distance between centres and the radius of the first circle. $R = \sqrt{AB^{2}+r^{2}} = \sqrt{(\sqrt{5})^{2}+2^{2}} = \sqrt{5+4} = \sqrt{9} = 3$.

If one of the diameters of the curve $x^{2}+y^{2}-4x-6y+9=0$ is a chord of a circle with centre $(1,1)$,the radius of this circle is

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