lf 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 A $3$
Given circle $x^{2} + y^{2} - 2 x - 6 y - 6 = 0$
Centre $C_{1}$ at $(1, 3)$
and radius $r_{1} = \sqrt{1 + 9 + 6} = 4$
Given, center $C_{2}$ at $(2, 1)$
$C_{1} C_{2} = \sqrt{5}$
Now, $r_{2}^{2} = 4 + 5 = 9$
$\Rightarrow r_{2} = 3$

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