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

\sqrt{3}

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\sqrt{2}

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3

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2

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Solution

The correct option is D $3$
$x^{2} + y^{2} - 2 x - 6 y + 6 = 0$
$(x - 1)^{2} + (y - 3)^{2} = 4$
Centre is (1, 3) and radius = 2
If r = radius of second circle then .
$r^{2} = 2^{2} + (2 - 1)^{2} + (1 - 3)^{2} = 9$
$r = 3$

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