The coordinates of the radical centre of three circles $x^{2} + y^{2} + 2 x + 3 y + 1 = 0, x^{2} + y^{2} + x + y + 2 = 0, x^{2} + y^{2} + 3 x + 2 y + 1 = 0$

(6, 30)

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(\frac{1}{3}, \frac{1}{3})

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(3, 0)

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None of these

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Solution

The correct option is B $(\frac{1}{3}, \frac{1}{3})$
$S_{1} : x^{2} + y^{2} + 2 x + 3 y + 1 = 0$
$S_{2} : x^{2} + y^{2} + x + y + 2 = 0$
$S_{3} : x^{2} + y^{2} + 3 x + 2 y + 1 = 0$
Radius axis of $S_{1}$ & $S_{2}$
$L_{1} : S_{1} - S_{2} = 0 \to x + 2 y - 1 = 0$
Radical axis of $S_{2}$ & $S_{3}$
$L_{2} : S_{2} - S_{3} = 0 \to - 2 x - y + 1 = 0$
Intersection of $L_{1}$ & $L_{2}$ is radical centre.
Radical centre is $(\frac{1}{3}, \frac{1}{3})$
$B$ is correct

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