The radical centre of the circles given by
$x^{2} + y^{2} - 8 x - 12 y = 1, x^{2} + y^{2} - 6 x - 4 y = 1, x^{2} + y^{2} - 5 x - 7 y = 1$ is

(4, 6)

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(3, 2)

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(\frac{5}{2}, \frac{7}{2})

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None of these

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Solution

The correct option is D None of these
If in the given equations of the circles constant terms are equal then radical centre lies on the point of intersection of $x$ and $y$ axis
$∴$ Radical centre $(0, 0)$

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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

x^{2} + y^{2} + 3 x + 2 y + 1 = 0

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

x^{2} + y^{2} + 5 x - 8 y + 15 = 0

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

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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

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