Radical centre of $x^{2} + y^{2} - x + 3 y - 3 = 0$ ,
$x^{2} + y^{2} - 2 x + 2 y + 2 = 0$ and $x^{2} + y^{2} + 2 x + 3 y - 9 = 0$ is

(2, 3)

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(3, 2)

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(- 2, 3)

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(- 3, - 2)

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Solution

The correct option is A $(2, 3)$
$s_{1} : x^{2} + y^{2} - x + 3 y - 3 = 0$
$s_{2} : x^{2} + y^{2} - 2 x + 2 y + 2 = 0$
$s_{3} : x^{2} + y^{2} + 2 x + 3 y - 9 = 0$
So, $e q^{n}$ of one radical axis is $s_{1} - s_{2} = 0$
$\Rightarrow x + y - 5 = 0 - - (1)$
& $e q^{n}$ of another radical axis is $s_{1} - s_{3} = 0$
$\Rightarrow 3 x - 6 = 0$
$x = 2 (2)$
from point of intersection of (1) & (2),
$x = 2$ and $y = 3$
$(2, 3)$ is radical centre of this given three circles.

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