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Power of a Point

The radical c...

Question

The radical centre of the circles $x^{2} + y^{2} = 1, x^{2} + y^{2} - 2 x - 3 = 0$ and $x^{2} + y^{2} - 2 y - 3 = 0$ is

(1, 1)

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(1, - 1)

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(- 1, - 1)

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(- 1, 1)

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Solution

The correct option is C $(- 1, - 1)$
The equations of the circles are:
$x^{2} + y^{2} = 1$ .... $(1)$
$x^{2} + y^{2} - 2 x - 3 = 0$ .... $(2)$
$x^{2} + y^{2} - 2 y - 3 = 0$ .... $(3)$
Equation $(1) -$ Equation $(2)$
$- 1 + 2 x + 3 = 0$
$x = - 1$
Equation $(1) -$ Equation $(3)$
$- 1 + 2 y + 3 = 0$
$y = - 1$
$∴$ Radical centre $(- 1, - 1)$

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