Coordinates of the centre of the circle which bisects the circumferences 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:

(- 3, - 3)

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(3, 3)

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(2, 2)

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(- 2, - 2)

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Solution

The correct option is A $(- 3, - 3)$
Let the equation of circle bisects the $3$ circles $x^{2} + y^{2} = 1$
$\begin{matrix} x^{2} + y^{2} + 2 x - 3 = 0 x^{2} + y^{2} + 2 y - 3 = 0 \end{matrix}$
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
$= g x + f y + c - 1 = 0$

Passes throw $(0, 0)$ i.e.
Center of $x^{2} + y^{2} = 1$
$C = 1$
$2 g x + f y + c - 2 x + 3 = 0$

Pass throw $(- 1, 0)$ i.e. center of $x^{2} + y^{2} + 2 x - 3 = 0$
$\begin{matrix} - 2 g + c + 5 = 0 \Rightarrow - 2 g + 6 = 0 \Rightarrow g = 3 2 g x + 2 f y + c - 2 y + 3 = 0 \end{matrix}$

Pass throw $(0, - 1)$
$\begin{matrix} - 2 f + c + 5 = 0 f = 3 x^{2} + y^{2} + 6 x + 6 y + 1 = 0 C = (- 3, - 3) \end{matrix}$

$∴$ Option $A$ is correct answer.

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Coordinates of the centre of the circle which bisects the circumferences of the circles x2+y2=1: x2+y2+2x−3=0 and x2+y2+2y−3=0 is:

Coordinates of the centre of the circle which bisects the circumferences 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: