The radius of the circle,having centre at (2, | vedclass

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(C) 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 centre of this cir

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$3$

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(C) 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 centre of this circle is $C(-g, -f) = (1, 3)$ and its radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-3)^2 - 6} = \sqrt{1 + 9 - 6} = \sqrt{4} = 2$.Let the required circle have centre $D(2, 1)$.Since a chord of the required circle is a diameter of the given circle,the length of this chord is equal to the diameter of the given circle,which is $2 	imes r = 2 	imes 2 = 4$.Let $AB$ be the chord of the required circle,which is a diameter of the given circle. The length of $AB = 4$,so $AC = CB = 2$.In the right-angled triangle $	riangle ACD$,$AD$ is the radius of the required circle $(R)$.$AD^2 = AC^2 + CD^2$.$CD$ is the distance between $D(2, 1)$ and $C(1, 3)$.$CD = \sqrt{(2 - 1)^2 + (1 - 3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.$R^2 = 2^2 + (\sqrt{5})^2 = 4 + 5 = 9$.$R = 3$.

The radius of the circle,having centre at $(2, 1)$ and one of its chords as a diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$,is

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