If $P_{1}, P_{2}, P_{3}$ are the perimeters of the three circles $x^{2} + y^{2} + 8 x - 6 y = 0, 4 x^{2} + 4 y^{2} - 4 x - 12 y - 186 = 0$ and $x^{2} + y^{2} - 6 x + 6 y - 9 = 0$ respectively

P_{1} < P_{2} < P_{3}

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P_{1} < P_{3} < P_{2}

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P_{3} < P_{2} < P_{1}

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P_{2} < P_{3} < P_{1}

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Solution

The correct option is A $P_{1} < P_{3} < P_{2}$
$x^{2} + y^{2} + 8 x - 6 y = 0$
$r_{1} = \sqrt{4^{2} + 3^{2}} = 5$
$4 x^{2} + 4 y^{2} - 4 x - 12 y - 186 = 0$
$x^{2} + y^{2} - x - 3 y - \frac{93}{2} = 0$
$r_{2} = \sqrt{{\frac{1}{2}}^{2} + {\frac{3}{2}}^{2} + \frac{93}{2}} = 7$
$x^{2} + y^{2} - 6 x + 6 y - 9 = 0$
$r_{3} = \sqrt{3^{2} + 3^{2} + 9} = 3 \sqrt{3}$
The perimeter is proportional to the radius. Hence,
$P_{2} > P_{3} > P_{1}$

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