The centres of the circles x-2 - y-2 - 1- x-2 - y-2 - 6x - 2y - 1 and x-2 - y-2 - 12x - 4y - 1-areNon-collinearSameCollinearNone of these-

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Obtaining Centre and Radius of a Circle from General Equation of a Circle

The centres o...

Question

The centres of the circles $x^{2} + y^{2} = 1, x^{2} + y^{2} + 6 x - 2 y = 1$ and $x^{2} + y^{2} - 12 x + 4 y = 1$ are

Same

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Collinear

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Non-collinear

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None of these

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Solution

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x^{2} + y^{2} = 1,

x^{2} + y^{2} + 6 x - 2 y = 1

x^{2} + y^{2} - 12 x + 4 y = 1

x^{2} + y^{2} = 1, x^{2} + y^{2} + 6 x - 2 y = 1

x^{2} + y^{2} - 12 x + 4 y = 1

x^{2} + y^{2} - 4 x - 6 y + 5 = 0

x^{2} + y^{2} - 2 x - 4 y - 1 = 0

x^{2} + y^{2} - 6 x - 2 y = 0

x^{2} + y^{2} - 4 x - 6 y + 5 = 0, x^{2} + y^{2} - 2 x - 4 y - 1 = 0, x^{2} + y^{2} - 6 x - 2 y = 0

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