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If one of the diameters of the circle, given the equation, $x^{2} + y^{2} - 4 x + 6 y - 12 = 0,$ is a chord of circle ”S”, whose centre is a $(- 3, 2),$ then the radius of “S” is

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$5$

$5 \sqrt{3}$

$5 \sqrt{2}$

$10$

answer is C.

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Detailed Solution

$C_{1} = (2, - 3), C_{2} (- 3, 2)$
$C_{1} C_{2} = 5 \sqrt{2}, r = 5,$
$r_{1} = \sqrt{{(C_{1} C_{2})}^{2} + r^{2}}$
$= \sqrt{50 + 25} = 5 \sqrt{3}$

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